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Let $a,b,c>0 : a+b+c=4.$ Find minimum $$T=\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}+\dfrac{ab}{a^2+b^2}+\frac{1}{a}+\frac{1}{b} +\frac{1}{c}.$$

For $a=b=c=\dfrac{4}{3},$ I got a value $\dfrac{15}{4}$

So we need to prove $$\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}+\dfrac{ab}{a^2+b^2}+\frac{1}{a}+\frac{1}{b} +\frac{1}{c}\ge \dfrac{15}{4}$$ After using C-S, $$\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}+\dfrac{ab}{a^2+b^2}\ge \frac{(ab+bc+ca)^2}{\sum_{cyc}ab(a^2+b^2)}$$ Id est, we will prove $$\frac{q^2}{q(16-2q)-4r}+\frac{q^2}{r}\ge \dfrac{15}{4}$$

How can I continue it ?

Hope some helps. Thank you.

Michael Rozenberg
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Dragon boy
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1 Answers1

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We'll prove that $\frac{15}{4}$ it's a minimal value.

Indeed, we need to prove that: $$\sum_{cyc}\frac{ab}{a^2+b^2}+\frac{1}{4}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq\frac{15}{4}$$ or $$\frac{1}{4}\left((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9\right)\geq\frac{3}{2}-\sum_{cyc}\frac{ab}{a^2+b^2}$$ or $$\sum_{cyc}\left(\frac{a}{b}+\frac{b}{a}-2\right)\geq2\sum_{cyc}\frac{(a-b)^2}{a^2+b^2}$$ or $$\sum_{cyc}(a-b)^2\left(\frac{1}{ab}-\frac{2}{a^2+b^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^4}{ab(a^2+b^2)}\geq0.$$

Michael Rozenberg
  • 194,933