<s>[INST] If $a+b+c+d=1$ so $\sum\limits_{cyc}\sqrt{a+b+c^2}\geq3$ Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=1$. Prove that:
$$\sqrt{a+b+c^2}+\sqrt{b+c+d^2}+\sqrt{c+d+a^2}+\sqrt{d+a+b^2}\geq3.$$
I tried C-S, Holder and more, but without success.
 [/INST] A computer solution using the Buffalo Way
After homogenization, it suffices to prove that, for all $a, b, c, d \ge 0$,
$$\sum_{\mathrm{cyc}} \sqrt{(a + b)(a+b+c+d) + c^2} \ge 3(a+b+c+d).$$
Denote
\begin{align}
&X = (a + b)(a+b+c+d) + c^2,\ Y = (b+c)(a+b+c+d) + d^2, \\
&Z = (c+d)(a+b+c+d) + a^2,\ W = (d+a)(a+b+c+d) + b^2.
\end{align}
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} + \sqrt{W} \ge 3(a+b+c+d)$.
WLOG, assume $d = \min(a, b, c, d)$. We split into four cases:

*

*$a\ge b \ge c \ge d$ or $a\ge c \ge b \ge d$ or $c\ge a\ge b \ge d$: By GM-HM, we have $\sqrt{X} + \sqrt{W} = \sqrt{X + W + 2\sqrt{XW}}\ge \sqrt{X + W + \frac{4XW}{X+W}}$
and $\sqrt{Y} + \sqrt{Z} \ge \sqrt{Y + Z + \frac{4YZ}{Y+Z}}$.
It suffices to prove that
$$\sqrt{X + W + \frac{4XW}{X+W}} + \sqrt{Y + Z + \frac{4YZ}{Y+Z}} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + W + \frac{4XW}{X+W}} \sqrt{Y + Z + \frac{4YZ}{Y+Z}}\\
\ge\ & 9(a+b+c+d)^2 - X - W - \frac{4XW}{X+W} - Y - Z - \frac{4YZ}{Y+Z}.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + W + \frac{4XW}{X+W}\right)\left(Y + Z + \frac{4YZ}{Y+Z}\right)\\
\ge \ & \left(9(a+b+c+d)^2 - X - W - \frac{4XW}{X+W} - Y - Z - \frac{4YZ}{Y+Z}\right)^2.
\end{align}
The Buffalo Way works.


*$b\ge c \ge a \ge d$: By GM-HM, we have
$\sqrt{X} + \sqrt{Y} \ge \sqrt{X + Y + \frac{4XY}{X+Y}}$ and $\sqrt{Z} + \sqrt{W} \ge \sqrt{Z + W + \frac{4ZW}{Z+W}}$.
It suffices to prove that
$$\sqrt{X + Y + \frac{4XY}{X+Y}} + \sqrt{Z + W + \frac{4ZW}{Z+W}} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + Y + \frac{4XY}{X+Y}} \sqrt{Z + W + \frac{4ZW}{Z+W}}\\
\ge\ & 9(a+b+c+d)^2 - X - Y - \frac{4XY}{X+Y} - Z - W - \frac{4ZW}{Z+W}.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + Y + \frac{4XY}{X+Y}\right)\left(Z + W + \frac{4ZW}{Z+W}\right)\\
\ge\ & \left(9(a+b+c+d)^2 - X - Y - \frac{4XY}{X+Y} - Z - W - \frac{4ZW}{Z+W}\right)^2.
\end{align}
The Buffalo Way works.


*$b \ge a \ge c \ge d$: By GM-HM, we have
\begin{align}
\sqrt{X} + \sqrt{Y} + \sqrt{W} &= \sqrt{X + Y + W + 2\sqrt{XY} + 2\sqrt{YW} + 2\sqrt{WX}}\\
&\ge \sqrt{X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}}.
\end{align}
It suffices to prove that
$$\sqrt{X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}} + \sqrt{Z} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}} \sqrt{Z}\\
\ge\ & 9(a+b+c+d)^2 - X - Y - W - \frac{4XY}{X+Y} - \frac{4YW}{Y+W} - \frac{4WX}{W+X} - Z.  
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}\right)Z\\
\ge\ & \left(9(a+b+c+d)^2 - X - Y - W - \frac{4XY}{X+Y} - \frac{4YW}{Y+W} - \frac{4WX}{W+X} - Z\right)^2.
\end{align}
The Buffalo Way works.


*$c \ge b \ge a \ge d$: By GM-HM, we have
\begin{align}
\sqrt{X} + \sqrt{Y} + \sqrt{Z} &= \sqrt{X + Y + Z + 2\sqrt{XY} + 2\sqrt{YZ} + 2\sqrt{ZX}}\\
&\ge \sqrt{X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}}.
\end{align}
It suffices to prove that
$$\sqrt{X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}} + \sqrt{W} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}}\sqrt{W}\\
\ge\ & 9(a+b+c+d)^2 - X - Y - Z - \frac{4XY}{X+Y} - \frac{4YZ}{Y+Z} - \frac{4ZX}{Z+X} - W.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}\right)W\\
\ge\ & \left(9(a+b+c+d)^2 - X - Y - Z - \frac{4XY}{X+Y} - \frac{4YZ}{Y+Z} - \frac{4ZX}{Z+X} - W\right)^2.
\end{align}
The Buffalo Way works.

We are done.
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