Hold on to faith

Algebra Level 3

If a , b a,b and c c are positive real numbers satisfying a b a 5 + b 5 + a b + c b b 5 + c 5 + c b + a c a 5 + c 5 + a c ≤ 1 \dfrac {ab}{a^5 +b^5+ab} + \dfrac{cb}{b^5+c^5+cb} + \dfrac {ac}{a^5+c^5+ac} \leq 1 , find a b c abc .


The answer is 1.

Surya Sharma by Surya Sharma , 20, India

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1 solution

Tom Engelsman
May 14, 2021

The leap-of-faith in question is to let a = b = c a=b=c which yields:

3 a 2 2 a 5 + a 2 = 3 2 a 3 + 1 ≤ 1 ⇒ 2 ≤ 2 a 3 ⇒ 1 ≤ a \Large \frac{3a^2}{2a^5+a^2} = \frac{3}{2a^3+1} \le 1 \Rightarrow 2 \le 2a^3 \Rightarrow 1 \le a

Thus, a b c = 1 3 = 1 . abc = 1^3 = \boxed{1}.

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