Inequality-1

Algebra Level pending

Let x , y , z > 0 x,y,z >0 and x + y + z = x y z x+y+z=xyz then:

1 x 2 + 1 + 1 y 2 + 1 + 1 z 2 + 1 \frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{z^2+1}}

has a maximum value of A B \dfrac{A}{B} , where A A and B B are coprime positive integers. What is A + B A+B ?


The answer is 5.

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1 solution

Hana Wehbi
Nov 12, 2018

First Solution: \underline{\textit{First Solution:}}

Set a = 1 x , b = 1 y , c = 1 z x y + y z + z x = 1 a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z} \implies xy+yz+zx=1 and the inequality: x x 2 + 1 3 2 \sum \frac{x}{\sqrt{x^2+1}}\le \frac{3}{2} ,

But x x 2 + 1 = x x 2 + x y + x z + z y = x x + y x x + z \sum \frac{x}{\sqrt{x^2+1}} = \sum \frac{x}{\sqrt{x^2+xy+xz+zy}} = \sum \sqrt{\frac{x}{x+y}\frac{x}{x+z}} ,

Also, x x + y x x + z x x + y + x x + z 2 \sum \sqrt{\frac{x}{x+y}\frac{x}{x+z}} \le \frac{\frac{x}{x+y}+\frac{x}{x+z}}{2} ,

Thus, x x + y x x + z x x + y + x x + z + y x + y + y y + z + z z + y + z z + x 2 = 3 2 \sum \sqrt{\frac{x}{x+y}\frac{x}{x+z}} \le \frac{\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y} +\frac{y}{y+z}+\frac{z}{z+y}+\frac{z}{z+x}}{2} = \frac{3}{2} ,

Second Solution: \underline{\textit{Second Solution:}}

Subsitute a = tan x , b = tan y , c = tan z where x + y + z = π a=\tan x, b=\tan y, c=\tan z \text{ where } x+y+z=\pi ,

then cos x + cos y + cos z 3 2 \cos x+\cos y+\cos z \le \frac{3}{2}

Third Solution: \underline{\textit{Third Solution:}}

By AM-GM we have a + b + c 3 a b c 3 and since a + b + c = a b c ( a b c ) 2 27 a+b+c \ge 3\sqrt[3]{abc} \text { and since } a+b+c = abc \implies (abc)^2 \ge 27 ,

We can rewrite the inequality as: 1 3 \frac{1}{3} ( 1 a 2 + 1 + 1 b 2 + 1 + 1 c 2 + 1 ) 1 2 (\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}) \le \frac{1}{2} ,

since f ( a ) = 1 a 2 + 1 f(a)=\frac{1}{\sqrt{a^2+1}} is concave, we apply Jensen's inequality:

1 3 f ( a ) + 1 3 f ( b ) + 1 3 f ( c ) f ( a + b + c 3 ) = f ( a b c 3 ) = 1 ( a b c ) 2 3 2 + 1 1 2 ( a b c ) 2 27. \frac{1}{3} f(a)+ \frac{1}{3} f(b) + \frac{1}{3} f(c) \le f( \frac{a+b+c}{3}) = f(\frac{abc}{3})= \frac{1}{\sqrt{\frac{(abc)^2}{3^2}+1}} \le \frac{1}{2} \iff (abc)^2\ge 27.

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