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

cauch

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}

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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