How prove $\sum\limits_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le ab+bc+ca$

Question

Let acute-angled triangle $ABC$,and $AB=c,BC=a,AC=b$,show that $$\sum_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le ab+bc+ca$$

I try use AM-GM $$\sum_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le\sum_{cyc}\dfrac{(a^2+b^2-c^2)+(a^2-b^2+c^2)}{2}=a^2+b^2+c^2$$ But this following not hold,because $$a^2+b^2+c^2\ge ab+bc+ac$$

Answer 1

Let $a^2+b^2-c^2=z^2$, $a^2+c^2-b^2=y^2$ and $b^2+c^2-a^2=x^2$, where $x$, $y$ and $z$ are positives.

Hence, we need to prove that $\sum\limits_{cyc}\sqrt{(x^2+y^2)(x^2+z^2)}\geq2(xy+cz+yz)$,

which after using C-S $\sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+yz$

gives $\sum\limits_{cyc}(x^2-xy)$, which is $\sum\limits_{cyc}(x-y)^2\geq0$.

Done!