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This question already has an answer here:

Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.

I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?

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marked as duplicate by user354271, kingW3, Frpzzd, Namaste algebra-precalculus Sep 2 '17 at 0:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Notice that by Inequality of arithmetic and geometric means we know that:
$$ 2ab \leq a^2+b^2; \\ 2ac \leq a^2+c^2; \\ 2bc \leq b^2+c^2; $$

so we can conclude that: $$ 2\left(ab+ac+bc\right) \leq 2\left(a^2+b^2+c^2\right) \Longrightarrow \\ \ \ \left(ab+ac+bc\right) \leq \ \ \left(a^2+b^2+c^2\right) \Longrightarrow \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \leq \ \ \left(a^2+b^2+c^2\right) . $$


Note that this in-equlality is sharp for $a=b=c=\sqrt{\dfrac{2}{3}}$;
for which one can see the value of $a^2+b^2+c^2$ is equal to $2$.

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Famke's answer is the simplest; however, we can use a variational argument, as well.

For all variations that maintain $ab+bc+ca=2$, we have $$ (b+c)\,\delta a+(a+c)\,\delta b+(a+b)\,\delta c=0\tag{1} $$ To minimize $a^2+b^2+c^2$, we must have $$ 2a\,\delta a+2b\,\delta b+2c\,\delta c=0\tag{2} $$ for all variations that satisfy $(1)$.

Orthogonality says that to have $(2)$ for all variations that satisfy $(1)$, we need $$ 2a=\lambda(b+c),\quad2b=\lambda(a+c),\quad\text{and}\quad2c=\lambda(a+b)\tag{3} $$ Summing these up, we get that $\lambda=1$, and then solving the equations, we get that $a=b=c$. Finally, to satisfy the constraint for $(2)$, we get that $a=b=c=\sqrt{\frac23}$. Thus, the minimum of $a^2+b^2+c^2$ is $2$.

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  • $\begingroup$ My dear robjohn♦ your solution is more creative and interesting. $\endgroup$ – Davood Sep 1 '17 at 12:46
  • $\begingroup$ Robjohn.Very elegant solution, Peter $\endgroup$ – Peter Szilas Sep 1 '17 at 14:21
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Scalar product:

Let $\vec A : = (a,b,c)$, $\vec B: = (b,c,a)$.

$|\vec A \cdot \vec B| \le |A| |B| $.

$\Rightarrow$ :

$2 \le ab +bc + ca \le a^2 +b^2 + c^2$.

Equality:

$a=b=c = \sqrt{\frac{2}{3}}$.

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You can solve this problem by your starting step.

Indeed, by C-S $$3(a^2+b^2+c^2)\geq(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc).$$ Thus, $$3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+ac+bc)$$ or $$a^2+b^2+c^2\geq ab+ac+bc$$ and since $ab+ac+bc\geq2,$ we obtain: $$a^2+b^2+c^2\geq2.$$ The equality occurs for $(1,1,1)||(a,b,c)$ and $ab+ac+bc=2$,

which says that $2$ is a minimal value.

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Again from Cauchy-Schwarz from a different perspective

$\sqrt{a^2c^2}+\sqrt{b^2c^2}+\sqrt{a^2c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$

$a^2+b^2+c^2 \ge ab+bc+ac \ge 2$

$a^2+b^2+c^2 \ge 2$

İnequality holds for $a=b=c=\sqrt{\dfrac{2}{3}}$

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