# Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ [duplicate]

Given that $a, b$ and $c$ are the sides of a triangle.

How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$? Maybe any hint? Am I going to wrong direction? $$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$ $$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$ $$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$ ...?

## marked as duplicate by Martin R, user1551, Frank, projectilemotion, user223391 Feb 21 '17 at 22:42

• There is a cube in the title of your question. – Zubzub Feb 21 '17 at 16:39
• @MichaelBurr, It is given that $a,b,c$ are sides of a triangle. – S.C.B. Feb 21 '17 at 16:47

Note that three sides $a,b,c$ in a triangle satisfy $$a<b+c, b<a+c, c<a+b.$$ So one has $$a^2<a(b+c), b^2<b(a+c), c^2<c(a+b).$$ Adding these three inequalities gives $$a^2+b^2+c^2<2ab+2bc+2ca.$$

Note that we have $$a^{2}>(b-c)^{2}\;,b^{2}>(a-c)^{2}\;,c^{2}>(b-a)^{2}$$ As $a,b,c$ are sides of a trinagle. Adding these, we get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$ Which is equivalent to $$a^2+b^2+c^2<2ab+2bc+2ca$$

You can even prove more: if $a,b,c$ are the sides and $\alpha,\beta,\gamma$ the respectively opposite angles, summing up the three identities of the law of cosine, we get $$a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ca\cos\beta$$ which implies the statement, as all the cosines are less than $1$.

Alternatively, by cosine law

\begin{align*} b^2+c^2-a^2 &= 2bc\cos A \\ c^2+a^2-b^2 &= 2ca\cos B \\ a^2+b^2-c^2 &= 2ab\cos C \\ a^2+b^2+c^2 &= 2(bc\cos A+ca\cos B+ab\cos C) \\ & \le 2(bc+ca+ab) \end{align*}

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x$, $y$ and $z$ are positives and $$\sum_{cyc}(2ab-a^2=\sum_{cyc}(2(x+z)(y+z)-(y+z)^2)=$$ $$=\sum_{cyc}(2x^2+6xy-2x^2-2xy)=4(xy+xz+yz)>0$$