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

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

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## Comments

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TopNewestExpanding 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: \[\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a+c}+\dfrac{a}{b+a}+\dfrac{b}{c+b}\\ =\dfrac{b^2}{ab}+\dfrac{c^2}{bc}+\dfrac{a^2}{ac}+\dfrac{c^2}{ac+c^2}+\dfrac{a^2}{ab+a^2}+\dfrac{b^2}{bc+b^2}\\ \geq \frac{(b+c+a+c+a+b)^2 }{2(ab+ac+bc)+a^2+b^2+c^2} =\frac{2^2(a+b+c)^2}{(a+b+c)^2}=\boxed{4}\]

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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.

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Oops, forgot. Thanks for telling

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First assume WLOG that a>=b>=c; now a+b>=a+c>=b+c; now when we take the inverses of both sequences; the order changes(1/a and 1/(a+b)) sequence. So apply rearrangement inequality to minimize the expression in second bracket (like club 1/c with 1/(a+b)). Write expression in first bracket as 1/2[a(b+c) + b(c+a) + c(a+b)] and now apply Cauchy Schwarz ineq. to get the result i.e. minimum value is 9/2.

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I think the answer is 4.5

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