Maximize a Constant in an Inequality with a Restriction

Algebra Level 4

Real numbers a a , b b , and c c satisfy a b c = 1 abc=1 . Given that k k is an integer 2 \geq 2 , find the maximum possible R R that satisfies

a k a + b + b k b + c + c k c + a R \dfrac{a^k}{a+b}+\dfrac{b^k}{b+c}+\dfrac{c^k}{c+a} \geq R

for all a a , b b , and c c satisfying the condition previously stated.


The answer is 1.5.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Advaith Kumar
May 29, 2020

notice that using Chebyshev, it is possible to generalise Nesbitt's inequality. consider the 2 increasing sequences a^n-1 >= b^n-1 >= c^n - 1 and a/b +c >=b/c+a>=a+b . Now use chebyshev and (nesbitt's later) and solve

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...