# Cauchy-Schwarz Inequality By Engr. Roy Roque Rivera Jr. Part 2

This is the second part of Cauchy – Schwarz Inequality By Engr. Roy Roque Rivera Jr. The primary application of this topic is proving hard inequality problems. This post is also recommended for PMO and IMO aspirants and other Mathematical Olympiads. Please also check the first part of this topic.

2.2 Proving Inequalities

Problem 1: Let $a,b,c \in \Re$. Prove that

$3(a^2+b^2+c^2)\geq (a+b+c)^2$

Solution: By using the Cauchy- Schwarz Inequality in Engel Form, we have

$a^2+b^2+c^2\geq \displaystyle\frac{(a+b+c)^2}{3}$

Hence,

$3(a^2+b^2+c^2)\geq (a+b+c)^2$

Problem 2 (Nesbitt’s Inequality – England, 1903): Let $a,b,c \in +\Re$. Prove that

$\displaystyle\frac{a}{b+c}+\displaystyle\frac{b}{c+a}+\displaystyle\frac{c}{a+b}\geq \displaystyle\frac{3}{2}$

Solution: The left-hand side of the inequality can be written as

$\displaystyle\frac{a^2}{ab+ac}+\displaystyle\frac{b^2}{bc+ab}+\displaystyle\frac{c^2}{ac+bc}\geq \displaystyle\frac{(a+b+c)^2}{2(ab+ac+bc)}$

Since $a^2+b^2+c^2\geq ab+ac+bc,$ we have

$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\geq ab+ac+bc+2(ab+ac+bc)$

Or

$(a+b+c)^2\geq 3(ab+ac+bc)$

Hence,

$\displaystyle\frac{a^2}{ab+ac}+\displaystyle\frac{b^2}{bc+ab}+\displaystyle\frac{c^2}{ac+bc}\geq \displaystyle\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq \displaystyle\frac{3(ab+ac+bc)}{2(ab+ac+bc)}$

Thus,

$\displaystyle\frac{a}{b+c}+\displaystyle\frac{b}{c+a}+\displaystyle\frac{c}{a+b}\geq \displaystyle\frac{3}{2}$

Problem 3 (Belarus IMO Team Selection Test, 1999): Let $a,b,c\in \Re$ such $a^2+b^2+c^2=3$. Prove that

$\displaystyle\frac{1}{1+ab}+\displaystyle\frac{1}{1+bc}+\displaystyle\frac{1}{1+ca}\geq \displaystyle\frac{3}{2}$

Solution: By using the Cauchy- Schwarz Inequality in Engel Form, we have

$\displaystyle\frac{1}{1+ab}+\displaystyle\frac{1}{1+bc}+\displaystyle\frac{1}{1+ca}\geq \displaystyle\frac{(1+1+1)^2}{(1+ab)+(1+bc)+(1+ca)}$

Or simply

$\displaystyle\frac{1}{1+ab}+\displaystyle\frac{1}{1+bc}+\displaystyle\frac{1}{1+ca}\geq \displaystyle\frac{9}{3+ab+ac+bc}$

But

$a^2+b^2+c^2\geq ab+ac+bc$

Because

$a^2+b^2+c^2\geq ab+ac+bc$

$2a^2+2b^2+2c^2\geq 2ab+2ac+2bc$

Which can be written as

$(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)\geq 0$

Or obviously

$(a-b)^2+(b-c)^2+(a-c)^2\geq 0$

Now,

$3+a^2+b^2+c^2\geq 3+ab+ac+bc\longrightarrow \displaystyle\frac{9}{3+a^2+b^2+c^2}\leq \displaystyle\frac{9}{3+a^2+b^2+c^2}$

Hence,

$\displaystyle\frac{a}{1+ab}+\displaystyle\frac{1}{1+bc}+\displaystyle\frac{1}{1+ca}\geq \displaystyle\frac{9}{3+ab+ac+bc}$

But we know that $a^2+b^2+c^2=3$; thus,

$\displaystyle\frac{a}{1+ab}+\displaystyle\frac{1}{1+bc}+\displaystyle\frac{1}{1+ca}\geq \displaystyle\frac{9}{3+3}$

Or simply

$\displaystyle\frac{a}{1+ab}+\displaystyle\frac{1}{1+bc}+\displaystyle\frac{1}{1+ca}\geq \displaystyle\frac{3}{2}$

Problem 4 (Ireland, 1999): Let $a,b,c,d \in +\Re$ such that $a+b+c+d=1$. Prove that

$\displaystyle\frac{a^2}{a+b}+\displaystyle\frac{b^2}{b+c}+\displaystyle\frac{c^2}{c+d}+\displaystyle\frac{d^2}{d+a}\geq \displaystyle\frac{1}{2}$

Solution: By using the Cauchy- Schwarz Inequality in Engel Form, we have

$\displaystyle\frac{a^2}{a+b}+\displaystyle\frac{b^2}{b+c}+\displaystyle\frac{c^2}{c+d}+\displaystyle\frac{d^2}{d+a}\geq \displaystyle\frac{(a+b+c+d)^2}{2(a+b+c+d)}$

But $a+b+c+d=1$; thus,

$\displaystyle\frac{a^2}{a+b}+\displaystyle\frac{b^2}{b+c}+\displaystyle\frac{c^2}{c+d}+\displaystyle\frac{d^2}{d+a}\geq \displaystyle\frac{1}{2}$

Problem 5 (Iran, 1999): Let $a,b,c>0$. Prove that

$(ab+bc+ac)(\displaystyle\frac{1}{(a+b)^2}+\displaystyle\frac{1}{(b+c)^2}+\displaystyle\frac{1}{(a+c)^2})\geq \displaystyle\frac{9}{4}$

Solution: By using the Cauchy – Schwarz Inequality in Engel Form, we have

$\displaystyle\frac{1}{(a+b)^2}+\displaystyle\frac{1}{(b+c)^2}+\displaystyle\frac{1}{(a+c)^2}\geq \displaystyle\frac{(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})^2}{3}$

Again by Cauchy-Schwarz Inequality in Engel Form

$\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{a+c}\geq \displaystyle\frac{3^2}{2(a+b+c)}$

Or

$\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{a+c}\geq \displaystyle\frac{9}{2(a+b+c)}$

It follows that

$\displaystyle\frac{1}{(a+b)^2}+\displaystyle\frac{1}{(b+c)^2}+\displaystyle\frac{1}{(a+c)^2}\geq \displaystyle\frac{(\frac{9}{2(a+b+c)})^2}{3}$

Or

$\displaystyle\frac{1}{(a+b)^2}+\displaystyle\frac{1}{(b+c)^2}+\displaystyle\frac{1}{(a+c)^2}\geq \displaystyle\frac{27}{4(a+b+c)^2}$

Now, since $a^2+b^2+c^2\geq ab+ac+bc$

$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\geq 3(ab+ac+bc)$

Hence,

$\displaystyle\frac{1}{(a+b)^2}+\displaystyle\frac{1}{(b+c)^2}+\displaystyle\frac{1}{(a+c)^2}\geq \displaystyle\frac{27}{4(a+b+c)^2}$

$\geq \displaystyle\frac{27}{4\cdot 3(ab+ac+bc)}$

$=\displaystyle\frac{9}{4(ab+ac+bc)}$

Or simply

$(ab+bc+ac)(\displaystyle\frac{1}{(a+b)^2}+\displaystyle\frac{1}{(b+c)^2}+\displaystyle\frac{1}{(a+c)^2})\geq \displaystyle\frac{9}{4}$

Problem 6: Let $x,y,z>0$. Prove that

$\sqrt{3x^2+xy}+\sqrt{3y^2+yz}+\sqrt{3z^2+xz}\geq 2(x+y+z)$

Solution: The left-hand side of the inequality can be written as

$\sqrt{x}\sqrt{3x+y}+\sqrt{y}\sqrt{3y+z}+\sqrt{z}\sqrt{3z+x}$

By using the Cauchy-Schwarz Inequality, we have

$\sqrt{x+y+z}\sqrt{(3x+y)+(3y+z)+(3z+x)}\geq \sqrt{x}\sqrt{3x+y}+\sqrt{y}\sqrt{3y+z}+\sqrt{z}\sqrt{3z+x}$

Or simply

$\sqrt{x+y+z}\sqrt{4x+4y+4z}\geq \sqrt{x}\sqrt{3x+y}+\sqrt{y}\sqrt{3y+z}+\sqrt{z}\sqrt{3z+x}$

Or

$2\sqrt{x+y+z}\sqrt{x+y+z}\geq \sqrt{x}\sqrt{3x+y}+\sqrt{y}\sqrt{3y+z}+\sqrt{z}\sqrt{3z+x}$

Hence,

$\sqrt{3x^2+xy}+\sqrt{3y^2+yz}+\sqrt{3z^2+zx}\leq 2(x+y+z)$

Problem 7 (International Mathematical Olympiad, 1995): Let $a,b,c>0$ such that $abc=1$. Prove that

$\displaystyle\frac{1}{a^3(b+c)}+\displaystyle\frac{1}{b^3(c+a)}+\displaystyle\frac{1}{c^3(a+b)}\geq \displaystyle\frac{3}{2}$

Solution: By Cauchy- Schwarz Inequality, we have

Or

$\displaystyle\frac{1}{a^3(b+c)}+ \displaystyle\frac{1}{b^3(c+a)}+ \displaystyle\frac{1}{c^3(a+b)}\geq \displaystyle\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{a(b+c)+b(c+a)+c(a+b)}$

Thus, it suffices to show that

$\displaystyle\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{a(b+c)+b(c+a)+c(a+b)}\geq \displaystyle\frac{3}{2}$

Since

$\displaystyle\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{a(b+c)+b(c+a)+c(a+b)}= \displaystyle\frac{ab+ac+bc}{2}$

From AM-GM Inequality, we have

$ab+ac+bc\geq 3\sqrt{a^2b^2c^2}$

But since $abc=1$; hence,

$ab+ac+bc\geq 3$

Which implies that

$\displaystyle\frac{ab+ac+bc}{2}=\displaystyle\frac{3}{2}$

It follows that

$\displaystyle\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{a(b+c)+b(c+a)+c(a+b)}\geq \displaystyle\frac{3}{2}$

And we are done.

Problem 8 (GDR, 1967): Prove that if $n\geq 2$ and $a_1,a_2,\ldots,a_n$ are positive real numbers whose sum is $S$, then

$\displaystyle\frac{a_1}{S-a_1}+\displaystyle\frac{a_2}{S-a_2}+\ldots+\displaystyle\frac{a_n}{S-a_n}\geq \displaystyle\frac{n}{n-1}$

Solution:  The left-hand side of the inequality can be written as

$\displaystyle\frac{a_1^2}{Sa_1-a_1^2}+\displaystyle\frac{a_2^2}{Sa_2-a_2^2}+\ldots+\displaystyle\frac{a_n}{Sa_n-a_n^2}$

Thus, by Cauchy-Schwarz Inequality in Engel Form, we have

$\displaystyle\frac{a_1^2}{Sa_1-a_1^2}+\displaystyle\frac{a_2^2}{Sa_2-a_2^2}+\ldots+\displaystyle\frac{a_n}{Sa_n-a_n^2}\geq \displaystyle\frac{(a_1+a_2+\ldots+a_n)^2}{S(a_1+a_2+\ldots+a_n)-(a_1^2+a_2^2+\ldots+a_n^2)}$

or

$\displaystyle\frac{a_1^2}{Sa_1-a_1^2}+\displaystyle\frac{a_2^2}{Sa_2-a_2^2}+\ldots+\displaystyle\frac{a_n}{Sa_n-a_n^2}\geq \displaystyle\frac{S^2}{S^2-( a_1^2+a_2^2+\ldots+a_n^2)}$

Recall the Quadratic Mean- Arithmetic Mean (QM-AM) Inequality

$\sqrt{\displaystyle\frac{a_{1}^2+a_{2}^2+\ldots+a_{n}^2}{n}}\geq \displaystyle\frac{a_1+a_2+\ldots+a_n}{n}$

Hence,

$\sqrt{\displaystyle\frac{a_{1}^2+a_{2}^2+\ldots+a_{n}^2}{n}}\geq \displaystyle\frac{S}{n}$

Or

$a_{1}^2+a_{2}^2+\ldots+a_{n}^2\geq \displaystyle\frac{S^2}{n}$

It follows that

$\displaystyle\frac{S^2}{S^2-(a_1^2+a_2^2+\ldots+a_n^2)}\geq \displaystyle\frac{S^2}{S^2(\frac{n-1}{n})}$

Hence,

$\displaystyle\frac{a_1}{S-a_1}+\displaystyle\frac{a_2}{S-a_2}+\ldots+\displaystyle\frac{a_n}{S-a_n}\geq \displaystyle\frac{n}{n-1}$

Problem 9 (Asia Pacific Mathematics Olympiad, 1990): Let $a_1,a_2,\ldots ,a_n$  and $b_1,b_2,\ldots ,b_n$  be positive real numbers such that

$a_1+a_2+\ldots+a_n=b_1+b_2+\ldots+b_n$

Prove that

$\displaystyle\frac{a_1^2}{a_1+b_1}+\displaystyle\frac{a_2^2}{a_2+b_2}+\ldots+\displaystyle\frac{a_n^2}{a_n+b_n}\geq \displaystyle\frac{a_1+a_2+\ldots+a_n}{2}$

Solution: From Cauchy – Schwarz Inequality in Engel Form, we have

$\displaystyle\frac{a_1^2}{a_1+b_1}+\displaystyle\frac{a_2^2}{a_2+b_2}+\ldots+\displaystyle\frac{a_n^2}{a_n+b_n}\geq \displaystyle\frac{(a_1+a_2+\ldots+a_n)^2}{a_1+a_2+\ldots+a_n+b_1+b_2+\ldots+b_n}$

But

$a_1+a_2+\ldots+a_n= b_1+b_2+\ldots+b_n$

Therefore,

$\displaystyle\frac{a_1^2}{a_1+b_1}+\displaystyle\frac{a_2^2}{a_2+b_2}+\ldots+\displaystyle\frac{a_n^2}{a_n+b_n}\geq \displaystyle\frac{(a_1+a_2+\ldots+a_n)^2}{2(a_1+a_2+\ldots+a_n)}$

Or simply

$\displaystyle\frac{a_1^2}{a_1+b_1}+\displaystyle\frac{a_2^2}{a_2+b_2}+\ldots+\displaystyle\frac{a_n^2}{a_n+b_n}\geq \displaystyle\frac{a_1+a_2+\ldots+a_n}{2}$

Problem 10 (Crux Mathematicorum): Let $x,y,z$  be positive real numbers. Prove that

$(\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}})^2+(\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}})^2+(\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2\geq 12$

Solution: From Cauchy- Schwarz Inequality in Engel Form, we have

$(\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}})^2+(\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}})^2+(\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2\geq \displaystyle\frac{(\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}}+\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}}+\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2}{3}$

It suffices to solve that

$(\displaystyle\frac{\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}}+\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}}+\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2}{3}\geq 12$

Now,

$(\displaystyle\frac{\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}}+\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}}+\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2}{3}=\displaystyle\frac{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x+y+z}{\root 3\of{xyz}})^2}{3}$

But, from AM-GM Inequality, we know that

$\displaystyle\frac{x}{y}+\displaystyle\frac{y}{z}+\displaystyle\frac{z}{x}\geq 3\root 3\of{\displaystyle\frac{x}{y}\cdot \displaystyle\frac{y}{z}\cdot \displaystyle\frac{z}{x}}$

Or simply

$\displaystyle\frac{x}{y}+\displaystyle\frac{y}{z}+\displaystyle\frac{z}{x}\geq 3$

Also,

$x+y+z=\root 3\of{xyz}\longleftrightarrow \displaystyle\frac{x+y+z}{\root 3\of{xyz}}\geq 3$

If follows that

$(\displaystyle\frac{\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}}+\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}}+\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2}{3}\geq \displaystyle\frac{(3+3)^2}{3}$

Hence,

$(\displaystyle\frac{\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}}+\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}}+\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2}{3}\geq 12$

Which implies that

$(\displaystyle\frac{x}{y}+\displaystyle\frac{z}{\root 3\of{xyz}})^2+(\displaystyle\frac{y}{z}+\displaystyle\frac{x}{\root 3\of{xyz}})^2+(\displaystyle\frac{z}{x}+\displaystyle\frac{y}{\root 3\of{xyz}})^2\geq 12$

Problem 11 (Phan Kim Hung): Let $a,b,c$ be positive real numbers. Prove that

$\displaystyle\frac{a^2-bc}{2a^2+b^2+c^2}+\displaystyle\frac{b^2-ca}{2b^2+c^2+a^2}+\displaystyle\frac{c^2-ab}{2c^2+a^2+b^2}\geq 0$

Solution: Note that

$\sum_{cyc} \displaystyle\frac{a^2-bc}{2a^2+b^2+c^2}\geq 0\longrightarrow -2\sum_{cyc} \displaystyle\frac{a^2-bc}{2a^2+b^2+c^2}\leq 0$

$\longleftrightarrow \sum_{cyc} \displaystyle\frac{-2a^2+2bc}{2a^2+b^2+c^2}\leq 0$

$\longleftrightarrow \sum_{cyc} (\displaystyle\frac{-2a^2+2bc}{2a^2+b^2+c^2}+1)\leq 3$

$\longleftrightarrow \sum_{cyc} \displaystyle\frac{b^2+2b+c^2c}{2a^2+b^2+c^2}\leq 3$

$\longleftrightarrow \sum_{cyc} \displaystyle\frac{(b+c)^2}{2a^2+b^2+c^2}\leq 3$

It suffices to show that

$\sum_{cyc} \displaystyle\frac{(b+c)^2}{2a^2+b^2+c^2}\leq 3$

But from Cauchy – Schwarz Inequality in Engel Form observe that

$\displaystyle\frac{b^2}{a^2+b^2}+\displaystyle\frac{c^2}{a^2+c^2}\geq \displaystyle\frac{(b+c)^2}{2a^2+b^2+c^2}$                                   $(\star)$

$\displaystyle\frac{c^2}{a^2+b^2}+\displaystyle\frac{a^2}{a^2+b^2}\geq \displaystyle\frac{(a+c)^2}{2b^2+a^2+c^2}$      $(\star\star)$

And

$\displaystyle\frac{a^2}{a^2+c^2}+\displaystyle\frac{b^2}{a^2+c^2}\geq \displaystyle\frac{(a+b)^2}{2c^2+a^2+b^2}$  $(\star\star\star)$

Thus, by adding $(\star), (\star\star), (\star\star\star)$ we obtain

$3\geq \displaystyle\frac{(b+c)^2}{2a^2+b^2+c^2}+\displaystyle\frac{(a+c)^2}{2b^2+a^2+c^2}+\displaystyle\frac{(a+b)^2}{2c^2+a^2+b^2}$

Or in shorthand notation we have

$\sum_{cyc} \displaystyle\frac{(b+c)^2}{2a^2+b^2+c^2}\leq 3$

Problem 12 (Iran, 1998): Let $x,y,z$  be positive real number such that $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=2$ Prove that

$\sqrt{x+y+z}\geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$

Solution: Observe

$\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=1\longleftrightarrow 3-(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z})=1$

Or equivalently

$\displaystyle\frac{x-1}{x}+\displaystyle\frac{y-1}{y}+\displaystyle\frac{z-1}{z}=1$

By Cauchy – Schwarz Inequality, we have

$((\displaystyle\frac{\sqrt{x-1}}{\sqrt{x}})^2+(\displaystyle\frac{\sqrt{y-1}}{\sqrt{y}})^2+(\displaystyle\frac{\sqrt{z-1}}{\sqrt{z}})^2((\sqrt{x})^2+(\sqrt{y})^2+(\sqrt{z})^2)\geq (\displaystyle\frac{\sqrt{x-1}}{\sqrt{x}}\cdot \sqrt{x}+\displaystyle\frac{\sqrt{y-1}}{\sqrt{y}}\cdot \sqrt{y}+\displaystyle\frac{\sqrt{z-1}}{\sqrt{z}}\cdot \sqrt{z})^2$

Or simply

$(\displaystyle\frac{x-1}{x}+ \displaystyle\frac{y-1}{y}+ \displaystyle\frac{z-1}{z})(x+y+z)\geq (\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$

Recalling that $\displaystyle\frac{x-1}{x}+\displaystyle\frac{y-1}{y}+\displaystyle\frac{z-1}{z}=1$, we obtain

$x+y+z\geq (\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$

It follows that

$\sqrt{x+y+z}\geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$