Question

Prove a desigualdade
${a}^{2} + b ^{2} + {c }^{2} \geqslant ab + bc + ca$
e determine quando ocorrer a igualdade​

Answer 1

$\displaystyle \sf a^2+b^2+c^2\geq ab+bc+ac \\\\ 2\cdot(a^2+b^2+c^2)\geq 2\cdot ( ab+bc+ac) \\\\ 2a^2+2b^2+2c^2\geq 2ab +2bc+2ac\\\\ 2a^2+2b^2+2c^2-2ab-2bc-2ac \geq 0 \\\\ \underbrace{\sf (a^2-2ab+b^2)}_{(a-b)^2} +\underbrace{\sf(a^2-2ac+c^2)}_{(a-c)^2}+\underbrace{\sf (b^2-2bc+c^2)}_{(b-c)^2 } \geq 0$

$\displaystyle \underbrace{\sf (a-b)^2}_{\displaystyle \sf \geq 0 }+\underbrace{\sf (a-c)^2}_{\displaystyle \sf \geq 0 }+\underbrace{\sf (b-c)^2}_{\displaystyle \sf \geq 0 }\geq 0 \\\\\\ \huge\boxed{\sf C.Q.D}\checkmark$  

$\displaystyle \sf \underline{\text{A igualdade ocorre quando a = b = c}}: \\\\ a^2+b^2+c^2\geq ab+bc+ac\\\\ a= b=c \\\\ \therefore \\\\ a^2+a^2+a^2\geq a.a+a.a+a.a \\\\ 3a^2\geq 3a^2 \\\\ 3a^2 = 3a^2 \\\\ a = a$