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 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 $$F\left(\dfrac 12,\dfrac13,\dfrac17,\dfrac{1}{42}\right)\approx3.056773\gt3\\ F\left(\dfrac {5}{12},\dfrac13,\dfrac17,\dfrac{1}{84}\right)\approx3.023735\gt3$$

  • $\begingroup$ I look for a solution, which we can write during a competition. $\endgroup$ – Michael Rozenberg Dec 22 '16 at 21:05
  • $\begingroup$ I hope someone can help you. I wanted but I could not. $\endgroup$ – Piquito Dec 22 '16 at 21:49

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