How to prove this 6-var inequality

Question

Let $a$, $b$, $c$, $x$, $y$, $z$ be real numbers and $ab+bc+ca>0$. Prove that\[\small\left(ab+bc+ca\right)\left(\frac{xy}{\left(a+c\right)\left(b+c\right)}+\frac{yz}{\left(a+b\right)\left(a+c\right)}+\frac{zx}{\left(b+c\right)\left(b+a\right)}\right)\le\frac{1}{4}\left(x+y+z\right)^2.\]

I'll state my current method in brief. Write the inequality as $x^2+mx+n\ge0$, a quadratic in $x$, where $m$ and $n$ are expressed in $y$, $z$, $a$, $b$, $c$. Actually the discriminant \[4n-m^2\overset{(1)}=16(acy+bcy-abz-acz)^2\cdot\frac{\sum\limits_{\rm cyc}^{a,b,c}ab}{\prod\limits_{\rm cyc}^{a,b,c}(a+b)^2}\ge0,\] which ends the proof.

But $(1)$ is barely feasible without a computer, I think. Thus, I thought about using Wolstenholme inequality, but cannot succeed.

I am here to ask for a better proof by hand.

Answer 1

The Wolstenholme inequality $$xy\cos\gamma+yz\cos\alpha+zx\cos\beta\le\frac12(x^2+y^2+z^2)\qquad(\alpha+\beta+\gamma=\pi)$$ is easily transformed to the form $$xy\cos^2\frac\gamma2+yz\cos^2\frac\alpha2+zx\cos^2\frac\beta2\le\frac14(x+y+z)^2$$ via the substitutions $\cos\gamma=2\cos^2\frac\gamma2-1$ and so on. To verify the original inequality is in this form, let $$\cos^2\frac\alpha2=\frac{ab+bc+ca}{(a+b)(a+c)},$$ $$\cos^2\frac\beta2=\frac{ab+bc+ca}{(b+c)(b+a)}.$$ Then we have $$\sin^2\frac\alpha2=\frac{a^2}{(a+b)(a+c)},$$ $$\sin^2\frac\beta2=\frac{b^2}{(b+c)(b+a)}.$$ Thus a little calculation yields $$\begin{align} \cos^2\frac\gamma2&=\sin^2\left(\frac\alpha2+\frac\beta2\right)\\ &=\left(\sin\frac\alpha2\cos\frac\beta2+\cos\frac\alpha2\sin\frac\beta2\right)^2\\ &=\frac{(ab+bc+ca)(a\pm b)^2}{(a+b)^2(b+c)(c+a)} \end{align} $$ But by symmetry we may assume $a,b$ have the same sign, hence $$\cos^2\frac\gamma2=\frac{ab+bc+ca}{(b+c)(c+a)},$$ which completes the proof (edit: or it reduces to proving the Wolstenholme inequality).

Answer 2

Your idea works very well!

Indeed, we need to prove that: $$x^2+y^2+z^2\geq pyz+qzx+rxy,$$ where $p=\frac{4(ab+ac+bc)}{(a+b)(a+c)}-2$,$q=\frac{4(ab+ac+bc)}{(a+b)(b+c)}-2$ and $r=\frac{4(ab+ac+bc)}{(a+c)(b+c)}-2$ or $$x^2-(qz+ry)x+y^2+z^2-pyz\geq0,$$ for which it's enough to prove that $$(qz+ry)^2-4(y^2+z^2-pyz)\leq0$$ or $$(4-r^2)y^2-2(2p+rq)yz+(4-q^2)z^2\geq0$$ and since $$4-r^2=(2-r)(2+r)=\left(4-\tfrac{4(ab+ac+bc)}{(a+c)(b+c)}\right)\cdot\tfrac{4(ab+ac+bc)}{(a+c)(b+c)}=\tfrac{16c^2(ab+ac+bc)}{(c^2+ab+ac+bc)^2}>0$$ (because $ab+ac+bc>0$ and we can assume $c^2>0$),it's enough to prove that: $$(2p+rq)^2-(4-r^2)(4-q^2)\leq0$$ or $$p^2+q^2+r^2+pqr\leq4$$ or $$\sum_{cyc}\left(\frac{4(ab+ac+bc)}{(a+b)(a+c)}-2\right)^2+\prod_{cyc}\left(\frac{4(ab+ac+bc)}{(a+b)(a+c)}-2\right)\leq4,$$ which is true because $$\sum_{cyc}\left(\frac{4(ab+ac+bc)}{(a+b)(a+c)}-2\right)^2+\prod_{cyc}\left(\frac{4(ab+ac+bc)}{(a+b)(a+c)}-2\right)=$$ $$=4\left(\sum_{cyc}\left(\frac{2(ab+ac+bc)}{(a+b)(a+c)}-1\right)^2+2\prod_{cyc}\left(\frac{2(ab+ac+bc)}{(a+b)(a+c)}-1\right)\right)=$$ $$=4\left(\sum_{cyc}\left(\tfrac{4(ab+ac+bc)^2}{(a+b)^2(a+c)^2}-\tfrac{4(ab+ac+bc)}{(a+b)(a+c)}+1-\tfrac{8(ab+ac+bc)^2}{(a+b)^2(a+c)(b+c)}+\tfrac{4(ab+ac+bc)}{(a+b)(a+c)}\right)+\tfrac{16(ab+ac+bc)^3}{\prod\limits_{cyc}(a+b)^2}-2\right)=$$ $$=4\left(\sum_{cyc}\left(\tfrac{4(ab+ac+bc)^2}{(a+b)^2(a+c)^2}-\tfrac{8(ab+ac+bc)^2}{(a+b)^2(a+c)(b+c)}\right)+\tfrac{16(ab+ac+bc)^3}{\prod\limits_{cyc}(a+b)^2}+1\right)=$$ $$=4\left(\frac{4(ab+ac+bc)^2}{\prod\limits_{cyc}(a+b)^2}\cdot\sum_{cyc}\left((b+c)^2-2(a+c)(b+c)+4ab\right)+1\right)=$$ $$=4\left(\frac{4(ab+ac+bc)^2}{\prod\limits_{cyc}(a+b)^2}\cdot\sum_{cyc}\left(2a^2+2ab-2a^2-6ab+4ab\right)+1\right)=4\leq4$$ and we are done!