Frederick Kramer

2022-07-12

Show that for real $a,b,c$, ${a}^{2}+{b}^{2}+{c}^{2}\ge ab+bc+ca$

Zackery Harvey

Expert

Let $f\left(a,b,c\right)={a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca$. Then we have, $f\left(ta,t,b,tc\right)={t}^{2}f\left(a,b,c\right)$. Hence, $f$ is homogeneous of degree two. For, $t\ne 0$, we have $f\left(a,b,c\right)>0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}f\left(ta,tb,tc\right)>0$. Therefore, we may make various normalizations. For example, we may set, $a=1,b=1+x,c=1+y$ and get ${x}^{2}+{y}^{2}-xy=\left(x-\frac{y}{2}{\right)}^{2}+\frac{3{y}^{2}}{4}>0$

malalawak44

Expert

Using caushy-Schwartz Inequality:
$\left({a}^{2}+{b}^{2}+{c}^{2}\right)\cdot \left({b}^{2}+{c}^{2}+{a}^{2}\right)\ge \left(ab+bc+ca{\right)}^{2}$
and equality hold when $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}.$

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