A beautiful generalization of the Nesbitt Inequality

by ayushkhaitan3437

I want to discuss a beautiful inequality, that is a generalization of the famous Nesbitt inequality:

(Romanian TST) For positive a,b,x,y,, prove that \frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\geq \frac{3}{a+b}

Clearly, if a=b, then we get Nesbitt’s inequality, which states that

\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq \frac{3}{2}.

This is question 14 on Mildorf’s “Olympiad Inequalities”, and its solution comprises finding a factor to multiply this expression with, almost out of thin air, and then use Cauchy Schwarz and AM-GM inequalities to prove the assertion. My solution is the following:

On interchanging a and b, the right hand side remains the same. However, the left hand side becomes

\frac{x}{az+by}+\frac{y}{ax+bz}+\frac{z}{ay+bx}\geq \frac{3}{a+b}

On adding these two inequalities, we get

\frac{x}{ay+bz}+\frac{x}{az+by}+\frac{y}{az+bx}+\frac{y}{ax+bz}+\frac{z}{ax+by}+\frac{z}{ay+bx}\geq \frac{6}{a+b}

Multiplying both sides by \frac{1}{2}(a+b) and then adding 6 on both sides, we get

2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})\geq 9

This is obviously true by Cauchy Schwarz. We will explain below how we got this expression.

Let us see what happens to \frac{x}{ay+bz}+\frac{x}{az+by} in some detail. After multiplying by \frac{1}{2}(a+b) and adding 2, we get

\frac{\frac{1}{2}(a+b)x+ay+bz}{ay+bz}+\frac{\frac{1}{2}(a+b)x+az+by}{az+by}\geq 2\frac{(a+b)(x+y+z)}{(a+b)(y+z)}=2\frac{(x+y+z)}{y+z}.

EDIT: I assumed that this was obviously true. However, it is slightly non-trivial that this is true. For \frac{a}{b}+\frac{c}{d}\geq 2\frac{a+c}{b+d}, the condition that should be true is that (b-d)(bc-ad)\geq 0. This is true in our case above.

After adding the other terms also, we get

2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})

As pointed above, this is clearly \geq 9 by Cauchy-Schwarz.

Hence proved

Note: For the sticklers saying this isn’t a rigorous proof, a rigorous proof would entail us assuming that

\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}< \frac{3}{a+b}, and then deriving a contradiction by proving that

2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})< 9, which is obviously false