Don't let you into chaos when you see a weird inequality

Algebra Level 4

$a\sqrt[3]{8+b-c}+b\sqrt[3]{8+c-a}+c\sqrt[3]{8+a-b}$

Let $a,b$ and $c$ be non-negative numbers satisfying $a^2+b^2+c^2=\dfrac{1}{3}$ .

Find the maximum value of the expression above.

Source: My school selection MO for grade 10.

The answer is 2.

1 solution

P C
Feb 21, 2016

Call the expression E, we set $A= a^2+b^2+c^2=\frac{1}{3}$ $S=8+b-c+8+c-a+8+a-b=24$ $Y=a+b+c$ Now applying Holder's Inequality we have $E^3\leq A.S.Y$ $\Leftrightarrow E^3\leq 8.Y$ Using Cauchy-Schwarz Inequality we get $Y\leq \sqrt{3(a^2+b^2+c^2)}=1$ , therefore $E\leq 2$ The equality holds when $a=b=c=\frac{1}{3}$

An easier way would be to see the symmetry in the equation $a\sqrt [ 3 ]{ 8+b-c } +b\sqrt [ 3 ]{ 8+c-a } +c\sqrt [ 3 ]{ 8+a-b }$ , and the fact that $a,b,c$ are restricted to a sphere. You conclude that this is at a maximum when $a=b=c$ , and thus $a=b=c=\frac{1}{3}$ .

Eiad Hamwi - 5 years, 3 months ago

But i think that does not work always. Even in symmetrical cases Its not necessary for equality to hold . but yes in most cases it does work

Prakhar Bindal - 5 years, 3 months ago

Yeah, that's probably true

Eiad Hamwi - 5 years, 3 months ago

When symmetry holds, the expression (to be optimised with a given set of constrains) is always a local extremum when all the variables are equal.

A Former Brilliant Member - 5 years, 3 months ago

Did the same way! :)

Prakhar Bindal - 5 years, 3 months ago

Don't let you into chaos when you see a weird inequality

Source: My school selection MO for grade 10.

The answer is 2.

1 solution

0 pending reports