$a^2 + b^2 + c^2 = 3$, prove that $a + b + c \ge ab + bc + ac$

Question

Let $a$, $b$ and $c$ be positive numbers such that $a^2 + b^2 + c^2 = 3$. Prove that: $$a + b + c \ge ab + bc + ac.$$

Using AM-GM inequality I proved, that $ab + bc + ac \le 3$

I have no idea about moving forward.

Any help is appreciated!

Answer 1

You have $$(a+b+c)^2 \geq 3(ab+bc+ca) = (a^2+b^2+c^2)(ab+bc+ca) \geq (ab+bc+ca)^2$$

So, $$a+b+c\geq ab+bc+ca.$$

Answer 2

Assume $a,b,c \ge 0\;$and$\;a^2 + b^2 + c^2 = 3$.

Let $e_1=a+b+c,\;$and let$\;e_2=ab+bc+ca$.

\begin{align*} \text{Then}\; &(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0\\[4pt] \implies\;&2(a^2+b^2+c^2) \ge 2(ab + bc + ca)\\[4pt] \implies\;&a^2 + b^2 + c^2 \ge ab + bc + ca\\[4pt] \implies\;&3 \ge ab + bc + ca\\[4pt] \implies\;&e_2 \le 3\\[8pt] \text{Identi}&\text{cally, $a^2+b^2 + c^2 = (a+b+c)^2 -2(ab + bc + ca)$, hence}\\[8pt] &e_1^2-2e_2 = 3\\[4pt] \implies\;&e_1^2 = 2e_2 + 3\\[4pt] \implies\;&e_1^2-e_2^2 = -e_2^2 + 2e_2 + 3\\[4pt] &\phantom{e_1^2-e_2^2} = (3-e_2)(1 + e_2)\\[4pt] &\phantom{e_1^2-e_2^2} \ge 0\\[2pt] \implies\;&e_1^2 \ge e_2^2\\[4pt] \implies\;&e_1 \ge e_2\\[4pt] &\text{as required.}\\[4pt] \end{align*}

Answer 3

By C-S $$a+b+c=(a+b+c)\sqrt{\frac{a^2+b^2+c^2}{3}}=$$ $$=(a+b+c)\sqrt{\frac{(1+1+1)(a^2+b^2+c^2)}{9}}\geq$$ $$\geq(a+b+c)\sqrt{\frac{(a+b+c)^2}{9}}=\frac{(a+b+c)^2}{3}\geq ab+ac+bc,$$ where the last inequality it's $$\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$ Done!