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For $a,b \in \mathbb R^+$ such that $a+b=1$, then:

$$\left(a+ \frac{1}{a}\right)^2+\left(b+ \frac{1}{b}\right)^2\ge \frac{25}{2}$$

Caleb Stanford
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Adán González
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  • Have you tried substituting $1-a$ for $b$ in the left side and expanding terms? Where did you get? I haven't done it, but it seems the obvious thing to try. – Ross Millikan Sep 18 '16 at 15:40
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    I suppose you mean $a,b>0$. The statement is not true with e.g. a=-0.5 and b=1.5 – H. H. Rugh Sep 18 '16 at 15:45
  • Since it's a famous problem, I am wondering if you have tried google. – Cave Johnson Sep 18 '16 at 15:47
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    @CaveJohnson: it's more fun here ... –  Sep 18 '16 at 15:48
  • It's a AM/GM issure. replace $a$ with $1/2 + k$ and $b$ with $1/2 -k$ and prove $((1/2 +k) + 1/(1/2+k))^2 + ((1/2 - k) + 1/(12 - k))^2 \ge (1/2 + 1/(1/2))^2 + (1/2 + 1/(1/2))^2 = (5/2)^2 + (5/2)^2 = 2*(25/4) = 25/2$. – fleablood Sep 18 '16 at 16:21

3 Answers3

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Since $x^2$ is convex then $$\left(a+ \frac{1}{a}\right)^2+\left(b+ \frac{1}{b}\right)^2\geq 2\left(\frac{a+ \frac{1}{a}+b+ \frac{1}{b}}{2}\right)^2=\frac{1}{2}\left(1+ \frac{1}{a}+ \frac{1}{b}\right)^2$$ So it suffices to show that $$\frac{1}{a}+ \frac{1}{b}\geq 4$$ which holds because $\frac{1}{a}+ \frac{1}{b}=\frac{a+b}{ab}=\frac{1}{a(1-a)}$.

Robert Z
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I suppose $a,b\geq 0$. By AM-QM we obtain $$\sqrt{\frac{\left(a+ \frac{1}{a}\right)^2+\left(b+\frac1b\right)^2}{2}}\geq \frac{\left(a+ \frac{1}{a}\right)+\left(b+\frac1b\right)}{2}=\frac{(a+b)+\left(\frac{a+b}{ab}\right)}{2}=\frac{1+\frac{1}{ab}}{2}.$$ By AM-GM $$ab\leq \left(\frac{a+b}{2}\right)^2=\frac14$$ therefore $$\frac{1+\frac{1}{ab}}{2}\geq \frac52$$ and taking the squares we conclude.

Del
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  • Just a quick tidbit about mathjax: instead of, e.g., $$(a+\frac 1a)^2$$ as you used, you can type \left(a +\frac 1a \right)^2 to get $$\left(a + \frac 1a \right)^2$$ – amWhy Sep 18 '16 at 15:51
  • @amWhy Yes, in fact I usually do. I'll edit thanks – Del Sep 18 '16 at 15:52
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expanding we get $$a^2+1/a^2+b^2+1/b^2\geq \frac{17}{2}$$ this is equivalent to $$a^2+b^2+\frac{a^2+b^2}{a^2b^2}\geq \frac{17}{2}$$ this is equivalent to $$(a^2+b^2)\left(1+\frac{1}{a^2b^2}\right)\geq \frac{17}{2}$$ from $a+b=1$ we get $a^2+b^2\geq 2ab$ and $$\frac{1}{ab}\geq 2$$ thus we get $$1+\frac{1}{a^2b^2}\geq 5$$ now you will need $$\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2=\frac{1}{4}$$

Dr. Sonnhard Graubner
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