Lockdown Math - Inequality

Algebra Level 2

If a , b , c > 0 a, b, c > 0 satisfying a b c = 1 , abc = 1, find the maximum value of a 2 a 3 + 1 + b 2 b 3 + 1 + c 2 c 3 + 1 \frac{a}{2a^3 + 1} + \frac{b}{2b^3 +1} +\frac{c}{2c^3 +1}


The answer is 1.

Adhiraj Dutta by Adhiraj Dutta , 18, India

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1 solution

The given expression is

S = a 2 a 3 + 1 + b 2 b 3 + 1 + c 2 c 3 + 1 = S=\dfrac{a}{2a^3+1}+\dfrac {b}{2b^3+1}+\dfrac{c}{2c^3+1}=

a 2 a 3 + 1 + b 2 b 3 + 1 + a 2 b 2 a 3 b 3 + 2 \dfrac{a}{2a^3+1}+\dfrac{b}{2b^3+1}+\dfrac{a^2b^2}{a^3b^3+2}

(since a b c = 1 abc=1 ).

This attains a maximum when

∂ S ∂ a = 0 , ∂ S ∂ b = 0 \dfrac{\partial S}{\partial a}=0, \dfrac{\partial S}{\partial b}=0 .

Differenting and doing some mathematics, this yields a − b = 0 ⟹ a = b a-b=0\implies a=b . The maximum value of the expression is

2 a 2 a 3 + 1 + a 4 a 6 + 2 \dfrac{2a}{2a^3+1}+\dfrac{a^4}{a^6+2} .

This will be maximum when it's rate of change goes zero, that is

2 − 8 a 3 ( 2 a 3 + 1 ) 2 + 8 a 3 − 2 a 9 ( a 6 + 2 ) 2 = 0 ⟹ a = 1 ⟹ b = a = 1 \dfrac{2-8a^3}{(2a^3+1)^2}+\dfrac{8a^3-2a^9}{(a^6+2)^2}=0\implies a=1\implies b=a=1 .

The maximum value is now

S m a x = 2 2 × 1 3 + 1 + 1 4 1 6 + 2 = 2 3 + 1 3 = 1 S_{max}=\dfrac{2}{2\times 1^3+1}+\dfrac{1^4}{1^6+2}=\dfrac{2}{3}+\dfrac{1}{3}=\boxed 1 .

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