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I can't find any solution for this inequality which can be found here

Exercise 1.3.5 Let $a,b,c$ be positive real numbers such that $a+b+c=ab+bc+ca$ and $n \leq 3$. Prove that

$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq 3+n$$

Thanks :)

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    $\begingroup$ Start by using Cauchy-Schwarz, $\displaystyle \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq \frac{(a+b+c)^2}{a+b+c}=a+b+c $. Also note that $(a+b+c)^2 \geq 3(ab+bc+ac) = 3(a+b+c)$ that further shows us that $(a+b+c) \geq 3$. Does it help? $\endgroup$ Commented Aug 31, 2012 at 10:57
  • $\begingroup$ Hello I am not getting How you go for $$(a+b+c)^{2}\geq 3(ab+bc+ac)$$ $\endgroup$
    – TerenceP
    Commented Nov 30, 2019 at 10:52

1 Answer 1

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Making use of given hint, $$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} = {(a + b +c)(a^2 + b^2 + c^2)\over (a+b+c)} = a^2 + b^2 +c^2$$ The minimum value of $a^2+b^2+c^2 \geq a+b+c \geq 3 $ for given criteria.

Let ${a^2 + b^2 + c^2 = z}$ then $f(z, n) = z + {3n \over z}$ for $n \leq 3$ and $z \geq 3$ $$f(z,n) = z + {3n \over z} \geq 6 \text{ for n=3}$$ For $n < 3$ the minimum value occurs at $f'(z) = 0 \implies z = \sqrt{3n } < 3 \leq z$ so further more $f'(z \geq 3) > 0$ suggest that $f(z)$ is increasing. So, the min-value must occur at $z=3$ which gives the desired result.

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