MRF2017 wrote: Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$. Prove that $$\sum_{cyc} \sqrt{\frac{a^2+1}{2}} \geq (3.\sum_{cyc} \sqrt{a}) -8$$

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MRF2017 wrote: Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$. Prove that $$\sum_{cyc} \sqrt{\frac{a^2+1}{2}} \geq (3.\sum_{cyc} \sqrt{a}) -8$$ $\sum_{cyc}\left(\sqrt{\frac{a^2+1}{2}}-3\sqrt{a}+2\right)=\sum_{cyc}\left(\sqrt{\frac{a^2+1}{2}}-3\sqrt{a}+2-4\left(\frac{1}{a+1}-\frac{1}{2}\right)\right)$. Let $a=x^2$, where $x>0$, and $x^2+1=2ux$. Hence, $u\geq1$ and it remains to prove that $\sqrt{\frac{x^4+1}{2}}-3x+2-4\left(\frac{1}{x^2+1}-\frac{1}{2}\right)\geq0$ or $(x^2+1)\sqrt{\frac{x^4+1}{2}}\geq x(3x^2-4x+3)$ or $u\sqrt{2u^2-1}\geq3u-2$ or $(u-1)^2(u^2+2u-2)\geq0$. Done!

Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4$. Prove that$$\sqrt[3]{\frac{a^3+b^3}{2}}+ \sqrt[3]{\frac{b^3+c^3}{2}} +\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3] {\frac{d^3+a^3}{2}} \leq 2(a+b+c+d)-4$$

It is a beautiful inequality. In fact the transformation is so skillful, not such easy to notice.

by changing variables $\frac{1}{a+1} \implies a=\frac{1-x}{x}$ and we define $y,z,t$ similarly and it implies $\sum x=2$ and now for the function $f=\sqrt{\frac{2x^2-2x+1}{2x^2}}-3\sqrt{\frac{1-x}{x}}$ is convex (didn't do it by hand, used geogebra ) so the minimum is obtained when all variable are equal