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Algebra Level 4

Positive real numbers a a , b b , and c c are such that 1 a + 1 b + 1 c = a + b + c \dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c} =a+b+c . And if

1 ( 2 a + b + c ) 2 + 1 ( a + 2 b + c ) 2 + 1 ( a + b + 2 c ) 2 ≤ p q \frac{1}{(2a+b+c)^2}+ \frac{1}{(a+2b+c)^2}+ \frac{1}{(a+b+2c)^2} \leq \dfrac{p}{q}

where p p and q q are coprime positive integers. Find p + q p+q .

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Atul Shivam by Atul Shivam , 23, India

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1 solution

Chew-Seong Cheong
Aug 12, 2016

Since the three terms of the LHS of the inequality are positive, we can apply RMS-AM inequality as follows:

1 3 ( 1 ( 2 a + b + c ) 2 + 1 ( a + 2 b + c ) 2 + 1 ( a + b + 2 c ) 2 ) ≤ 1 3 ( 1 ( 2 a + b + c ) 4 + 1 ( a + 2 b + c ) 4 + 1 ( a + b + 2 c ) 4 ) ⟹ 1 ( 2 a + b + c ) 2 + 1 ( a + 2 b + c ) 2 + 1 ( a + b + 2 c ) 2 ≤ 3 ( 1 ( 2 a + b + c ) 4 + 1 ( a + 2 b + c ) 4 + 1 ( a + b + 2 c ) 4 ) \begin{aligned} \frac 13 \left(\frac 1{(2a+b+c)^2} + \frac 1{(a+2b+c)^2} + \frac 1{(a+b+2c)^2}\right) & \le \sqrt{\frac 13 \left( \frac 1{(2a+b+c)^4} + \frac 1{(a+2b+c)^4} + \frac 1{(a+b+2c)^4}\right)} \\ \implies \frac 1{(2a+b+c)^2} + \frac 1{(a+2b+c)^2} + \frac 1{(a+b+2c)^2} & \le\sqrt{3 \left( \frac 1{(2a+b+c)^4} + \frac 1{(a+2b+c)^4} + \frac 1{(a+b+2c)^4}\right)} \end{aligned}

Equality occurs when a = b = c a=b=c , ⟹ 1 a + 1 b + 1 c = a + b + c \implies \dfrac 1a + \dfrac 1b + \dfrac 1c = a + b + c , ⟹ 3 a = 3 a \implies \dfrac 3a = 3a , ⟹ a = b = c = 1 \implies a = b = c = 1 .

∴ 1 ( 2 a + b + c ) 2 + 1 ( a + 2 b + c ) 2 + 1 ( a + b + 2 c ) 2 ≤ 3 ( 3 4 4 ) = 3 16 \begin{aligned} \therefore \frac 1{(2a+b+c)^2} + \frac 1{(a+2b+c)^2} + \frac 1{(a+b+2c)^2} & \le \sqrt{3 \left( \frac 3{4^4} \right)} = \frac 3{16} \end{aligned}

⟹ p + q = 3 + 16 = 19 \implies p + q = 3 + 16 = \boxed{19}

5 Helpful 1 Interesting 1 Brilliant 0 Confused

Good solution sir did exactly the same

A Former Brilliant Member - 3 years, 4 months ago

I don't follow the logic of this solution. You've shown that A<=B, with equality iff a=b=c, in which case A=B=3/16. But why can't there be another value of (a,b,c) where A, B are both larger and A<B?

Joe Mansley - 3 months ago

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Difficult to explain but in this case there is a maximum for A A for only a triple ( a , b , c ) (a,b,c) when A = B A=B . Any other values of ( a , b , c ) (a,b,c) , A < B A<B . You can try with cases other than a = b = c = 1 a=b=c=1 and you will find that A < 3 16 A < \frac 3{16} .

Chew-Seong Cheong - 3 months ago

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