Show that: $a^4+b^4+c^4<2a^2b^2+2b^2c^2+2a^2c^2$

Question

If $a,b,c$ are triangle sides, show that:

$$a^4+b^4+c^4<2a^2b^2+2b^2c^2+2a^2c^2$$

It's very similar to $a^2+b^2+c^2≥ab+bc+ac \Rightarrow a^4+b^4+c^4>a^2b^2+b^2c^2+a^2c^2$

I can't continue from here.

Answer 1

Note that the sides of a triangle must satisfy \begin{eqnarray*} a+b-c>0 \\ a-b+c >0 \\ -a+b+c >0 . \end{eqnarray*} So \begin{eqnarray*} (a+b+c) (a+b-c)(a-b+c)(-a+b+c)>0. \end{eqnarray*} Expand this & you will have your inequality.

It is subtly different from \begin{eqnarray*} (a-b)^2 +(b-c)^2+(c-a)^2 >0 \\ a^2+b^2+c^2 > ab+bc+ca. \end{eqnarray*}

Answer 2

HİNT:

I've solved this before.

$$(a+b+c)×(a+b-c)×(b+c-a)×(a+c-b)>0$$

And you will get the required inequality.

Answer 3

The difference betwen LHS and RHS can be written under the form of a determinant:

$$\tag{1}\Delta=\begin{vmatrix}0 &a^2&b^2&1\\a^2&0&c^2&1\\b^2&c^2&0&1\\1&1&1&0\end{vmatrix} = a^4 +b^4+c^4 − 2(a^2b^2+a^2c^2+b^2c^2) $$

$\Delta$ is known as the Cayley-Menger determinant.

Why is $\Delta \leq 0$ ? Because there is a relationship between $\Delta$ and the area $A$ of the triangle (which is a kind of re-writing of the famous Heron's formula):

$$\tag{2}A^2=\dfrac{-1}{2^2(2!)^2} \Delta$$

(see pages 9-10 of the excellent document (https://arxiv.org/pdf/1502.02816.pdf).

Remark 1: Due to (2), the given inequality cannot become an equality unless $A=0$, i.e., unless the triangle is flat.

Remark 2: Relationship (1) considered as a quadratic form in $a^2, b^2, c^2$ has rather rich applications. See for example "About the Neuberg-Pedoe and the Oppenheim Inequalities, D. S. Mitrinovic and J. E. Pecaric, Journal of Math. Anal. and Appl. 129, 196-210 (1988)"

Answer 4

$$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=4a^2b^2-(a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2)=$$ $$(2ab)^2-(a^2+b^2-c^2)^2=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)=$$ $$=(c^2-(a-b)^2)((a+b)^2-c^2)=(c-a+b)(c+a-b)(a+b-c)(a+b+c)>0.$$