Suppose $a^2+b^2+c^2<2(ab+bc+ca)$ $(a-b-c)^2<4bc$ $-2\sqrt{bc}<a-b-c<2\sqrt{bc}$ $(\sqrt{b}-\sqrt{c})^2<a<(\sqrt{b}+\sqrt{c})^2$ $|\sqrt{b}-\sqrt{c}|<\sqrt{a}<\sqrt{b}+\sqrt{c}$ so $\sqrt{a},\sqrt{b},\sqrt{c}$ are sides of a triangle. WLOG assume $x\ge y\ge z$. We have $x(y-z)^2+z(x-y)^2-y(z-x)^2=(x-y)(z-y)(x+z)\le0$, so $\sqrt{a}+\sqrt{c}\le\sqrt{b}$, a contradiction.

lssl wrote: Let $x,y,z $ be positive reals such that $\sqrt{a}=x(y-z)^2$, $\sqrt{b}=y(z-x)^2$ and $\sqrt{c}=z(x-y)^2$. Prove that \[a^2+b^2+c^2 \geq 2(ab+bc+ca)\] note that: $a^2 + b^2 + c^2 - 2\left( {ab + bc + ca} \right) = - \left( {\sqrt a + \sqrt b + \sqrt c } \right)\prod\limits_{cyc} {\left( {\sqrt a + \sqrt b - \sqrt c } \right)} $ but $\sqrt a + \sqrt b - \sqrt c = x \left( {y - z} \right)^2 + y\left( {z - x} \right)^2 - z\left( {x - y} \right)^2$ $= - \left( {x + y} \right)\left( {y - z} \right)\left( {z - x} \right)$ so $a^2 + b^2 + c^2 - 2\left( {ab + bc + ca} \right) = - \left( {\sqrt a + \sqrt b + \sqrt c } \right)\prod\limits_{cyc} {\left( { - \left( {x + y} \right)\left( {y - z} \right)\left( {z - x} \right)} \right)} $ $ = \left( {\sqrt a + \sqrt b + \sqrt c } \right)\prod\limits_{cyc} {\left( {\left( {x + y} \right)\left( {x - y} \right)^2 } \right)} \ge 0$ done.

In fact, the inequality is equivalent to $((a-b)(b-c)(c-a))^2(a+b)(b+c)(c+a)(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-6abc)\ge0$.

Let $x,y,z\geq 0, a=x^2(y-z)^4,b=y^2(z-x)^4,c=z^2(x-y)^4$. Prove that$$\left( a+b+c \right)^3\geq\frac{9}{2}\left( a\sqrt{b}+b\sqrt{c}+c\sqrt{a} \right)^2$$2021 Beijing University Golden Autumn Camp D2 P1 7587155041