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Algebra Level 5

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

Given that a , b a,b and c c are positive reals satisfying a 2 b 2 + b 2 c 2 + a 2 c 2 ≤ 666 a 2 b 2 c 2 a^2b^2+b^2c^2+a^2c^2\leq666a^2b^2c^2 . Let P P denote the maximum value of the expression above.

Find 10 P 10P , round to the nearest integer.

Part of the set


The answer is 112.

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P C by P C , 20, Vietnam

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1 solution

P C
Jan 30, 2016

First, we simplify J to J = 1 2 a + b + c + 1 2 b + a + c + 1 2 c + a + b J=\frac{1}{2a+b+c}+\frac{1}{2b+a+c}+\frac{1}{2c+a+b} then, prove J ≤ 1 4 ( 1 a + 1 b + 1 c ) J\leq\frac{1}{4}\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big) using AM-HM inequality

Then, using Cauchy - Schwartz inequality 1 4 ( 1 a + 1 b + 1 c ) ≤ 1 4 3 ( 1 a 2 + 1 b 2 + 1 c 2 ) \frac{1}{4}\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\leq\frac{1}{4}\sqrt{3\bigg(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)} From the condition we get 1 a 2 + 1 b 2 + 1 c 2 ≤ 666 \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\leq666 J ≤ 3.666 4 = p ≈ 11.17 J\leq\frac{\sqrt{3.666}}{4}=p\approx11.17 10 p = 111.7 10p = 111.7 which is 112 when rounded to the nearest integer

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Wow the same way!!

Department 8 - 5 years, 4 months ago

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Shit I thought we had to find greatest integer less than or equal to 10P ; thus answered 111.My mistake :P

Harsh Shrivastava - 5 years, 4 months ago

Can you please explain the second step, ​please?

Naman Singh - 5 years, 4 months ago

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We just applying AM-HM 1 2 a + b + c = 1 ( a + b ) + ( a + c ) ≤ 1 4 ( 1 a + c + 1 a + b ) \frac{1}{2a+b+c}=\frac{1}{(a+b)+(a+c)}\leq\frac{1}{4}(\frac{1}{a+c}+\frac{1}{a+b}) Similarly we have 1 2 b + a + c ≤ 1 4 ( 1 a + b + 1 b + c ) \frac{1}{2b+a+c}\leq\frac{1}{4}(\frac{1}{a+b}+\frac{1}{b+c}) and 1 2 c + a + b ≤ 1 4 ( 1 a + c + 1 b + c ) \frac{1}{2c+a+b}\leq\frac{1}{4}(\frac{1}{a+c}+\frac{1}{b+c})

Combining them and we get J ≤ 1 2 ( 1 a + b + 1 a + c + 1 b + c ) J\leq\frac{1}{2}(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}) Clearly see that 1 a + b ≤ 1 4 ( 1 a + 1 b ) \frac{1}{a+b}\leq\frac{1}{4}(\frac{1}{a}+\frac{1}{b}) Therefore J ≤ 1 4 ( 1 a + 1 b + 1 c ) J\leq\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})

P C - 5 years, 4 months ago

maximum is attained when a=b=c

Srikanth Tupurani - 1 year, 8 months ago

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