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$\left ( ab+bc+ca \right )\left[\frac{1}{a(a+c)}+\frac{1}{b(b+a)}+\frac{1}{c(c+b)}\right]$

Let a,b,c are reals possitive number. Find the minimum of the expression above.

Note by Ms Ht
8 months, 3 weeks ago

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Expanding and simplifying,we get: $\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a+c}+\dfrac{a}{b+a}+\dfrac{b}{c+b}$ Now applying titu's lemma ,we get: $\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a+c}+\dfrac{a}{b+a}+\dfrac{b}{c+b}\\ =\dfrac{b^2}{ab}+\dfrac{c^2}{bc}+\dfrac{a^2}{ac}+\dfrac{c^2}{ac+c^2}+\dfrac{a^2}{ab+a^2}+\dfrac{b^2}{bc+b^2}\\ \geq \frac{(b+c+a+c+a+b)^2 }{2(ab+ac+bc)+a^2+b^2+c^2} =\frac{2^2(a+b+c)^2}{(a+b+c)^2}=\boxed{4}$ · 8 months, 2 weeks ago

While you did prove that

$$(ab+bc+ca)(\frac{1}{a(a+c)}+\frac{1}{b(b+a)}+\frac{1}{c(c+b)})\geq4$$

you did not find the minimum of the expression, since 4 is not a possible value. You can see that 4 is not possible if you look at the requirements for equality when using Titu's Lemma. · 8 months ago

Oops, forgot. Thanks for telling · 8 months ago