If $a+b+c+d=1$ so $\sum\limits_{cyc}\sqrt{a+b+c^2}\geq3$

Question

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.

For example, we need to prove that: $$\sum_{cyc}\sqrt{a^2+(c+d)(a+b+c+d)}\geq3(a+b+c+d).$$ Now, by Holder $$\left(\sum_{cyc}\sqrt{a^2+(c+d)(a+b+c+d)}\right)^2\sum_{cyc}\tfrac{(a+c+d)^3}{a^2+(c+d)(a+b+c+d)}\geq27(a+b+c+d)^3$$ and it's enough to prove $$3(a+b+c+d)\geq\sum_{cyc}\frac{(a+c+d)^3}{a^2+(c+d)(a+b+c+d)},$$ but for $c=d=0$ and $a=b=1$ we obtain: $$3\cdot2\geq1+\frac{8}{3}+2+\frac{1}{2}$$ or$$6\geq6\frac{1}{6},$$ which is wrong, but almost true.

Maybe we can choose numbers $k>0$ and $m>0$ such that $k+m=2$ and the inequality $$3(a+b+c+d)\geq\sum_{cyc}\frac{(a+kc+md)^3}{a^2+(c+d)(a+b+c+d)}$$ would be true, but I have no an enough software to do this.

Answer 1

(For this difficult question that has not yet received an answer I will use the method of Lagrange multipliers).

Let $F$ be defined by $$F(x,y,z,w)=\sqrt{x+y+z^2}+\sqrt{y+z+w^2}+\sqrt{z+w+x^2}+\sqrt{w+x+y^2}$$ Using Lagrange multipliers we can find out the minimum of $F$ with the restriction $$x+y+z+w=1$$ We have the equation$$\nabla F(x,y,z,w)=\lambda\nabla(x+y+z+w)\qquad (1)$$ where $\nabla$ is the well known differential operator nabla. Put for short $$\begin{cases}D_1=\sqrt{z+w+x^2}\\D_2=\sqrt{w+x+y^2}\\D_3=\sqrt{x+y+z^2}\\D_4=\sqrt{y+z+w^2}\end{cases}$$

From $(1)$ we have $$\begin{cases}\dfrac{1}{2D_3}+\dfrac{1}{2D_2}+\dfrac{2x}{2D_1}=\lambda\\\dfrac{1}{2D_3}+\dfrac{1}{2D_4}+\dfrac{2y}{2D_2}=\lambda\\\dfrac{1}{2D_4}+\dfrac{1}{2D_1}+\dfrac{2z}{2D_3}=\lambda\\\dfrac{1}{2D_1}+\dfrac{1}{2D_2}+\dfrac{2w}{2D_4}=\lambda\end{cases}\qquad (2)$$ The system $(2)$ is clearly satisfied by $x=y=z=w$ which gives the values $\dfrac 14$ for each of the coordinates and $\dfrac 53$ for the Lagrange multiplier $\lambda$.

It follows $D_1=D_2=D_3=D_4=\dfrac 34$ and the required minimum is in fact equal to $4\cdot\dfrac 34=3$.

To see rigourously that $x=y=z=w=\dfrac 14$ gives a minimum for $F$ we have to verified that the limited Hessian matrix of the auxiliary function $h(x,y,z,w)=F(x,y,z,w)-\lambda(x+y+z+w)$ defined by (we have put $-1$ instead of the first partial derivatives of $x+y+z+w$) $$H=\begin{pmatrix}0&-1&-1&-1&-1\\-1&\dfrac{\partial^2 h}{\partial x^2}&\dfrac{\partial^2 h}{\partial x\partial y}&\dfrac{\partial^2 h}{\partial x\partial z}&\dfrac{\partial^2 h}{\partial x\partial w}\\-1&\dfrac{\partial^2 h}{\partial x\partial y}&\dfrac{\partial^2 h}{\partial y^2}&\dfrac{\partial^2 h}{\partial y\partial z}&\dfrac{\partial^2 h}{\partial y\partial w}\\-1&\dfrac{\partial^2 h}{\partial y^2}&\dfrac{\partial^2 h}{\partial y\partial z}&\dfrac{\partial^2 h}{\partial z^2}&\dfrac{\partial^2 h}{\partial \partial w}\\-1&\dfrac{\partial^2 h}{\partial x\partial w}&\dfrac{\partial^2 h}{\partial y\partial w}&\dfrac{\partial^2 h}{\partial z\partial w}&\dfrac{\partial^2 h}{ \partial w^2}\end{pmatrix}$$ is such that for the values $x=y=z=w=\dfrac 14$ all the determinants of the submatrices on the diagonal of order $3$ or greater than $3$ are positive. This is easy but something tedious so I do not do such a check and I just see it does not correspond to a maximum because, for example, close to $(0.25,0.25,0.25,0.25)$ one has $$F(0.251,0.252,0.248,0.249)\approx3.000003252\gt3$$ and for other two examples $$F\left(\dfrac 12,\dfrac13,\dfrac17,\dfrac{1}{42}\right)\approx3.056773\gt3\\ F\left(\dfrac {5}{12},\dfrac13,\dfrac17,\dfrac{1}{84}\right)\approx3.023735\gt3$$

Answer 2

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:

1. $a\ge b \ge c \ge d$ 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.

2. $b\ge c \ge a \ge d$: 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.

3. $b \ge a \ge c \ge d$: 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.

4. $c \ge b \ge a \ge d$: 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.