Given three positive numbers $x,y,z$. Prove that $(x+y-z)(\frac{3}{x+y}-\frac{1}{y+z}-\frac{1}{z+x})\leqq\frac{1}{2}$ .

Question

Given three positive numbers $x, y, z$. Prove that $$(x+ y- z)\left ( \frac{3}{x+ y}- \frac{1}{y+ z}- \frac{1}{z+ x} \right )\leqq \frac{1}{2}$$

First solution. Subtracting $(\!x+ y- z\!)(\!\frac{3}{x+ y}- \frac{1}{y+ z}- \frac{1}{z+ x}\!)$ from $1\div 2$, we have the following one: $$\frac{(x+ 2y)(z+ x- 2y)^{2}+ 2(4x+ 5y+ 18z)(x+ y- 2z)^{2}+ 30y(y- z)^{2}+ 39x(x- z)^{2}}{48(x+ y)(y+ z)(z+ x)}\geqq 0$$ How about another solution? I hope to see that. Thanks!

Answer 1

Hint
If $x+y\leqslant z$ the inequality is obvious*. Else we may write it as $$\frac3{x+y}\leqslant \frac1{x+z}+\frac1{y+z}+\frac1{2x+2y-2z}$$

which easily follows from Jensen’s inequality.

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PS: In case not so obvious, $$\frac1{x+z}+\frac1{y+z} \leqslant \frac1{2x+y} + \frac1{x+2y} \leqslant \frac2{x+y}$$

Answer 2

We need to prove that $f(z)\geq0$, where $$f(z)=6z^3-3(x+y)z^2-3(x^2+y^2)z+2x^3+x^2y+xy^2+2y^3.$$ Now, $$f'(z)=3(6z^2-2(x+y)z-x^2-y^2),$$ which gives $$z_{min}=\frac{x+y+\sqrt{7x^2+7y^2+2xy}}{6}$$ and we need to prove that $$f\left(z_{min}\right)\geq0,$$ which is smooth.

Indeed, $f\left(z_{min}\right)\geq0$ gives $$2(13x^3+3x^2y+3xy^2+13y^3)\geq\sqrt{(7x^2+7y^2+2xy)^3}.$$ Let $x^2+y^2=2kxy.$

Hence, $k\geq1$ and we need to prove that $$4(x+y)^2(13x^2+13y^2-10xy)^2\geq(7x^2+7y^2+2xy)^3$$ or $$4(k+1)(13k-5)^2\geq(7k+1)^3$$ or $$(k-1)(37k^2+38k-11)\geq0,$$ which is obvious.