$$3\geq ab+bc+ca\geq \dfrac{1}{3}$$

Find the range of values that can be taken by :

1) $\color{white}{s} a+b+c$

2) $\color{white}{s}abc$

For 1, I did: $$a^2+b^2+c^2\geq ab +bc+ca$$ $$a^2+b^2+c^2+ 2ab +2bc+2ca \geq 3ab +3bc+3ca $$ $$(a+b+c)^2\geq 3(ab+bc+ca)$$ Then I used the condition given to put the minimum value:$$(a+b+c)^2\geq 3(\dfrac{1}{3})$$ And I got $$(a+b+c) \geq(1)$$

Is this correct? And how do I get the maximum value?

  • $\begingroup$ From $(a+b+c)^2\geq1$ we get $a+b+c\geq1$ or $a+b+c\leq-1$. The maximum value of $a+b+c$ does not exist of course. $\endgroup$ – Michael Rozenberg Sep 20 '16 at 4:47

Use $A.M. \geq G.M. \geq H.M.$ on $ab,bc,ca$ to get

$\frac{ab + bc + ca}{3} \geq (ab.bc.ca)^{1/3} \geq \frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}$

$\Rightarrow \frac{ab + bc + ca}{3} \geq (abc)^{2/3} \geq \frac{3(abc)^2}{a+b+c}$

Similarly you can apply the same inequalities on $a,b,c$ to get

$\frac{a + b + c}{3} \geq (abc)^{1/3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$\Rightarrow \frac{a + b + c}{3} \geq (abc)^{1/3} \geq \frac{3(abc)}{ab+bc+ca}$

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  • $\begingroup$ I think you are wrong! $a$, $b$ and $c$ can be negatives. $\endgroup$ – Michael Rozenberg Sep 20 '16 at 5:22

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