Given four real numbers $a,b,c,d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that $a^2+b^2+c^2+d^2\leq ab+ac+ad+bc+bd+cd.$ 
Given four real numbers $a, b, c, d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that
$$a^{2}+ b^{2}+ c^{2}+ d^{2}\leq ab+ ac+ ad+ bc+ bd+ cd$$

My solution
$$3a- d\geq 0$$
$$\begin{align}\Rightarrow d\left ( a+ b+ c \right )- d^{2}= d\left ( a+ b+ c- d \right ) & = d\left ( 3a- d \right )+ d\left ( \left ( b- a \right )+ \left ( c- a \right ) \right )\\ 
 & \geq b\left ( b- a \right )+ c\left ( c- a \right ) \\ 
 & \geq \left ( b- a \right )^{2}+ \left ( c- a \right )^{2} \\ 
 & \geq \frac{1}{2}\left ( \left ( b- a \right )^{2}+ \left ( c- a \right )^{2}+ \left ( c- a \right )^{2} \right )\\
 & \geq \frac{1}{2}\left ( \left ( b- a \right )^{2}+ \left ( c- b \right )^{2}+ \left ( c- a \right )^{2} \right )\\
 & = a^{2}+ b^{2}+ c^{2}- ab- bc- ca
\end{align}$$
How about you ?

It's wrong.
Try $$(a,b,c,d)=(1,1,1,4).$$
For these values we need to prove that $$19\leq15,$$ which is not so true.
The following inequality is true already.

let $\{a,b,c,d\}\subset[1,3].$ Prove that:
$$a^2+b^2+c^2+d^2\leq ab+ac+bc+ad+bd+cd.$$

We can prove this inequality by the Convexity.
Indeed, let $f(a)=ab+ac+bc+ad+bd+cd-a^2-b^2-c^2-d^2$.
Thus, $f$ is a concave function, which says that $f$ gets a minimal value for an extreme value of $a$,
id est, for $a\in\{1,3\}$.
Similarly, for $b$, $c$ and $d$.
Thus, it's enough to check our inequality for $\{a,b,c,d\}\subset\{1,3\}$, which gives that our inequality is true.
