Given four edge-lengths of a quadrangle $a,b,c,d$ so that $a\leq b\leq c\leq d$. Prove that $a^{2}+b^{2}+c^{2}+d^{2}<2\left(ab+ac+ad+bc+bd+cd\right).$

Question

Given four edge-lengths of a quadrangle $a, b, c, d$ so that $a\leq b\leq c\leq d$. Prove that $$a^{2}+ b^{2}+ c^{2}+ d^{2}< 2\left ( ab+ ac+ ad+ bc+ bd+ cd \right )$$

My solution $$a+ b+ c> d$$ $$\begin{align}\Rightarrow \left ( 2a+ 2b+ 2c \right )d+ \left ( 2a+ 2b \right )c+ 2ab & > 2\left ( a+ b+ c \right )d+ ac+ bc\\ & > 2d^{2}+ a^{2}+ b^{2}\\ & \geq c^{2}+ d^{2}+ a^{2}+ b^{2} \end{align}$$ How about you ?

Answer 1

\begin{align}\Rightarrow 2\left ( ab+ ac+ ad+ bc+ bd+ cd \right ) & = 2\left ( a+ b+ c \right )d+ ab+bc+bc+ca+ca+ab\\ & \gt 2d^{2}+ \left (c+ a \right)b + \left (a+ b \right)c + \left (b+ c \right)a\\ & \gt 2d^{2} + b^{2} + \left (a + b \right)c + a^{2} \\ & \gt c^2 + d^{2} + b^{2} + \left (a + b \right)c + a^{2} \\ & \gt a^2 + b^{2} + c^{2} + d^{2} \end{align}