If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$

My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$ Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$.
Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!


marked as duplicate by Martin Sleziak, Aqua, Joel Reyes Noche, Xander Henderson, jvdhooft Oct 10 '17 at 13:14

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  • $\begingroup$ What techniques do you know/are you supposed to know/use? Do you know Lagrange multipliers? $\endgroup$ – TMM Jan 13 '14 at 12:59
  • $\begingroup$ No sir, I am not aware of that, but I know about QM-AM-GM-HM,Cauchy-Schwarz, and Chebyshev inequality. $\endgroup$ – Hawk Jan 13 '14 at 13:01
  • $\begingroup$ $\dfrac{1}{a^2}=\dfrac{a+b}{a^2}$.You can try to use this and see if you can get anything out of it. $\endgroup$ – rah4927 Jan 13 '14 at 13:07
  • $\begingroup$ Another suggestion-$a^2+b^2=(a+b)^2-2ab=1-2ab$. $\endgroup$ – rah4927 Jan 13 '14 at 13:10
  • $\begingroup$ Does the second suggestion help at all? Because I tried to make something out of it but ended up getting stuck! $\endgroup$ – Hawk Jan 13 '14 at 13:12

By QM-AM, $\displaystyle a^2+b^2 \geq \frac{1}{2}(a+b)^2 = \frac{1}{2}$. QM-AM again gives $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2$. By AM-HM, $\displaystyle \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \frac{a+b}{2} = \frac{1}{2}$, whence $\displaystyle\frac{1}{a}+\frac{1}{b} \geq 4$, and thus $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq 8$. Thus we have $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 = a^2 + b^2 + \frac{1}{a^2}+\frac{1}{b^2} + 4 \geq \frac{1}{2} + 8 + 4 = \frac{25}{2}.$$

  • $\begingroup$ Pretty Fine. Thank you! I tried this approach but I mixed up the equation and got stuck. Thanks for riddling this out! $\endgroup$ – Hawk Jan 13 '14 at 13:14

Hint: Substitute $a=\frac{1}{2}+x$ and $b=\frac{1}{2}-x$, $|x|<\frac{1}{2}$. Then you should only find the minimum of a one variable function $f(x)$.

I'll write the whole solution, maybe someone finds it helpful. After the substitution we get:


Since $x^2\geqslant0$, we get that $f(x)\geqslant\frac{9}{2}+\frac{\frac{1}{2}}{\frac{1}{16}}=\frac{25}{2}$. Equality: for $x=0$ or $a=b=\frac{1}{2}$.

  • $\begingroup$ The expressions only become uglier by that transformation, so then you could just as well use $b = 1 - a$ and optimize in $a$. $\endgroup$ – TMM Jan 13 '14 at 13:12
  • $\begingroup$ Breaking the symmetry by insisting $0 \leq x < \frac{1}{2}$ could very well simplify things. And the fact that the variables are rewritten symmetrically can help too -- e.g. the common denominator will be a completed square without extra work. $\endgroup$ – Hurkyl Jan 13 '14 at 13:18

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