Phorphyrion wrote: Let $x,y,z$ be non-negative real numbers. Prove that: \[\sqrt{(2x+y)(2x+z)}+\sqrt{(2y+x)(2y+z)}+\sqrt{(2z+x)(2z+y)}\geq \]\[\geq \sqrt{(x+2y)(x+2z)}+\sqrt{(y+2x)(y+2z)}+\sqrt{(z+2x)(z+2y)}.\]

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Let $x,y,z$ be non-negative real numbers. Prove that \[\sqrt{(3x+y)(3x+z)}+\sqrt{(3y+x)(3y+z)}+\sqrt{(3z+x)(3z+y)} \]\[\geq \sqrt{(x+3y)(x+3z)}+\sqrt{(y+3x)(y+3z)}+\sqrt{(z+3x)(z+3y)}\]

Let $x,y,z$ be non-negative real numbers. Prove that \[\sqrt{(kx+y)(kx+z)}+\sqrt{(ky+x)(ky+z)}+\sqrt{(kz+x)(kz+y)} \]\[\geq \sqrt{(x+ky)(x+kz)}+\sqrt{(y+kx)(y+kz)}+\sqrt{(z+kx)(z+ky)}\]Where $k>1.$

Phorphyrion wrote: Let $x,y,z$ be non-negative real numbers. Prove that: \[\sqrt{(2x+y)(2x+z)}+\sqrt{(2y+x)(2y+z)}+\sqrt{(2z+x)(2z+y)}\geq \]\[\geq \sqrt{(x+2y)(x+2z)}+\sqrt{(y+2x)(y+2z)}+\sqrt{(z+2x)(z+2y)}.\] Since the desired inequality is symmetric, assume that $x \ge y \ge z$. Let $a = (2x+y)(2x+z), b = (2y+x)(2y+z), c = (2z+x)(2z+y)$ and $A = (x + 2y)(x+2z), B = (y + 2x)(y + 2z), C = (z+2x)(z+2y)$. Clearly, we have $a \ge b \ge c$. We have $A - B = (x-y)(x+y-2z) \ge 0$ and $C - B = (y-z)(2x-y-z)\ge 0$. We have $abc = ABC$. We have $B - c = (2z+y)(x+y-2z) \ge 0$. We have $a - A = 3x^2 - 3yz \ge 0$ and $a - C = (2x+z)(2x-y-z) \ge 0$. Let $u_1 = \ln a, u_2 = \ln b, u_3 = \ln c$ and $v_1 = \ln A, v_2 = \ln B, v_3 = \ln C$. Then $(v_1, v_2, v_3)$ is majorized by $(u_1, u_2, u_3)$. By Karamata's inequality, we have $$\mathrm{e}^{u_1/2} + \mathrm{e}^{u_2/2} + \mathrm{e}^{u_3/2} \ge \mathrm{e}^{v_1/2} + \mathrm{e}^{v_2/2} + \mathrm{e}^{v_3/2}.$$ We are done.

sqing wrote: Let $x,y,z$ be non-negative real numbers. Prove that \[\sqrt{(kx+y)(kx+z)}+\sqrt{(ky+x)(ky+z)}+\sqrt{(kz+x)(kz+y)} \]\[\geq \sqrt{(x+ky)(x+kz)}+\sqrt{(y+kx)(y+kz)}+\sqrt{(z+kx)(z+ky)}\]Where $k>1.$ Karamata's inequality here

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