Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.

Question

Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.

That's what I've tried:

Let a Cauchy-Schwarz Inequality be :

\begin{array} (((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2 )\left(\left(\cfrac{ab}{\sqrt{a}}\right)^2 +\left(\cfrac{ac} {\sqrt{b}}\right)^2+\left(\cfrac{bc}{\sqrt{c}}\right)^2\right)& \ge (ab+ac+bc)^2 \\ (a+b+c)(ab^2+\cfrac{a^2 c^2}{b}+b^2 c) &\ge (ab+ac+bc)^2 \\ \end{array}

However how should I now factorize the left hand side of the inequality as $(a^2+b^2+c^2) \cdot Something $ ?

I think I've made the problem harder than it needs to be .

Answer 1

By C-S $\sum\limits_{cyc}(1+1)(a^2+b^2)\geq\sum\limits_{cyc}(a+b)^2$ and we are done!

Answer 2

If you insist on C-S, let $$ u=(a,b,c),\quad v=(b,c,a). $$ Then, $$ \text{RHS}=u\cdot v\underbrace{\leq}_{\text{C-S}}||u||\times||v||=\text{LHS}. $$

Answer 3

I have given a detailed proof using an easier method. Please click on this link to see the proof