Find the least value of $\dfrac {3a}{b+c} + \dfrac{4b}{a+c}+ \dfrac{5c}{a+b}$ for positive $a, b, c$.

I tried using the Cauchy-Schwarz inequality, but could not proceed after a bad equation which couldn't be further solved .

  • $\begingroup$ Are there rules about limiting a,b, and c in this to a number system? Otherwise, I'd suggest considering $a=b=c$ with as great a negative value as possible. $\endgroup$ – JB King Sep 28 '14 at 6:17
  • $\begingroup$ It is give that a,b,c are real positive numbers. $\endgroup$ – user167045 Sep 28 '14 at 6:32

You can minimize it by using Cauchy-Schwarz, but yes, it gets pretty bad. Firstly you can modify the expression as follows: $$\frac {3a}{b+c} + \frac{4b}{c+a}+ \frac{5c}{a+b}=\frac {3(a+b+c)}{b+c} + \frac{4(a+b+c)}{c+a}+ \frac{5(a+b+c)}{a+b}-12=(a+b+c)\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)-12=\frac{1}{2}\cdot ((b+c)+(c+a)+(a+b))\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)-12$$ Now, by Cauchy-Schwarz we have: $$((b+c)+(c+a)+(a+b))\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)\ge(\sqrt3+\sqrt4+\sqrt5)^2$$ With equality if: $$\frac{b+c}{\sqrt3}=\frac{c+a}{\sqrt4}=\frac{a+b}{\sqrt5}$$ So, if we use the modified expression from above, we get: $$\frac {3a}{b+c} + \frac{4b}{c+a}+ \frac{5c}{a+b}\ge\frac{1}{2}\cdot(\sqrt3+\sqrt4+\sqrt5)^2-12=\sqrt{12}+\sqrt{15}+\sqrt{20}-6\approx 5.809$$ To find a triplet $(a,b,c)$ for wich equality holds, the system of equations from above can be, after big effort, simplified to: $$a=\frac{\sqrt5+\sqrt4-\sqrt3}{\sqrt5-\sqrt4+\sqrt3}\cdot b=\frac{\sqrt4+\sqrt5-\sqrt3}{\sqrt4-\sqrt5+\sqrt3}\cdot c$$ And indeed, if we set $b=\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a$ and $c=\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a$ in the original expression, we get: $$\frac {3\cdot a}{\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a+\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a} + \frac{4\cdot\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a}{\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a+a}+ \frac{5\cdot\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a}{a+\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a}=\frac{1}{2}\cdot (\sqrt3 \cdot(\sqrt5+\sqrt4-\sqrt3)+\sqrt4 \cdot(\sqrt5-\sqrt4+\sqrt3)+\sqrt5 \cdot(-\sqrt5+\sqrt4+\sqrt3))=\sqrt{12}+\sqrt{15}+\sqrt{20}-6$$

So we're done.

I'm sorry for all the mistakes in my English, I'm not a native speaker.

  • $\begingroup$ Nice +1. You don't need to find the triplet, its mere existence guarantees that you have found the minimum. $\endgroup$ – Macavity Sep 29 '14 at 5:19

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