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@above $x-y, y-z, z-x$ can be negative.

$(x-y)(\sqrt{3x^2+y^2}-(x+y))=\frac{2x(x-y)^2}{\sqrt{3x^2+y^2}+x+y} \geq 0$ so $(x-y)\sqrt{3x^2+y^2} \geq x^2-y^2$ And so $(x-y)\sqrt{3x^2+y^2}+(y-z)\sqrt{3y^2+z^2}+(z-x)\sqrt{3z^2+x^2} \geq 0$

RagvaloD wrote: $(x-y)(\sqrt{3x^2+y^2}-(x+y))=\frac{2x(x-y)^2}{\sqrt{3x^2+y^2}+x+y} \geq 0$ so $(x-y)\sqrt{3x^2+y^2} \geq x^2-y^2$ And so $(x-y)\sqrt{3x^2+y^2}+(y-z)\sqrt{3y^2+z^2}+(z-x)\sqrt{3z^2+x^2} \geq 0$ nice ideas!

Let $x, y, z>0.$ Prove that$$(x-y)\sqrt{2x^2+y^2}+(y-z)\sqrt{2y^2+z^2}+(z-x)\sqrt{2z^2+x^2} \geq 0$$

a_507_bc wrote: Prove that for all positive reals $x, y, z$, the inequality $(x-y)\sqrt{3x^2+y^2}+(y-z)\sqrt{3y^2+z^2}+(z-x)\sqrt{3z^2+x^2} \geq 0$ is satisfied. WE EMPLOY THE PROOF BY EXAMPLE WE SET x = y = z = 1. THE THING BECOMES 0 ≥ 0, which is true !!!!!!!!!!!!!!!!!!!!!!

can't we just set $z = 1$ and do multivariable calculus thingies on the resulting expression

megarnie wrote: a_507_bc wrote: Prove that for all positive reals $x, y, z$, the inequality $(x-y)\sqrt{3x^2+y^2}+(y-z)\sqrt{3y^2+z^2}+(z-x)\sqrt{3z^2+x^2} \geq 0$ is satisfied. WE EMPLOY THE PROOF BY EXAMPLE WE SET x = y = z = 1. THE THING BECOMES 0 ≥ 0, which is true !!!!!!!!!!!!!!!!!!!!!! We need to prove the problem which satisfying for all positive real x,y,z not just x=y=z=1 case.

This is equivalent to show that $$\sum x \sqrt{3x^2 + y^2} \geq \sum y \sqrt{3x^2 + y^2}$$By $\text{AM-GM}$ we have $$\sum x \sqrt{3x^2 + y^2} \geq \sum x(\frac{3}{2}x + \frac{1}{2}y) = \frac{3}{2} \sum x^2 + \frac{1}{2}\sum xy.$$ On the other hand, by Cauchy–Schwarz, we can write $$\sum y \sqrt{3x^2 + y^2} = \sum xy \sqrt{3 + \frac{y}{x}} \leq \sqrt{\left( \sum x^2y^2 \right) \left( \sum 3 + \frac{y}{x} \right)}.$$ So, it suffices to show that $$ \left( \frac{3}{2} \sum x^2 + \frac{1}{2}\sum xy \right)^2 \geq \left( \sum x^2y^2 \right) \left( \sum 3 + \frac{y}{x} \right)$$ After some simple use of $\text{AM-GM}$ and removing similar terms, it is equivalent to $$\sum x^2y^2 + \sum xy^3 \geq 2xyx \sum x.$$ By considering $xy = a$, $yz = b$ and $zx = c$, the above inequality is equivalent to $$\sum a^2 + \sum \frac{a^2b}{c} \geq 2 \sum ab.$$And this is easily proved by $\text{AM-GM}$ as follow: $$\sum a^2 + \sum \frac{a^2b}{c} \geq \sum bc + \sum \frac{a^2b}{c} \geq 2 \sum ab.$$ And we are done.$\blacksquare$