If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge12.5$$

I could expand everything: $$a^2+2+\frac{1}{a^2}+b^2+2+\frac{1}{b^2}\ge12.5$$

Subtract $4$ from both sides: $$a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}\ge8.5$$

We could use AM-GM here ($a^2,\frac{1}{a^2},b^2,\frac{1}{b^2}$ are all positive), but obviously it wouldn't do anything useful. And we still have to use the fact that $a+b=4$ somehow.

I've tried substituting $b$ with $4-a$, but after clearing the denominators and simplifying we don't quite see anything useful, just a random 6th degree polynomial.

The polynomial is actually: $$2a^6-24 a^5+103.5 a^4-188 a^3+122 a^2-8 a+16\ge 0$$

How could I solve this? We can't use calculus by the way. Thanks.

  • $\begingroup$ The polynomial has real roots $2$. $\endgroup$
    – Don Larynx
    Dec 23, 2013 at 17:35
  • $\begingroup$ You may let $f(a,b)$ be your function and use your favourite method of optimization in two variables to show that at the global minimum , the value of the function is greater than or equal to $12.5$. $\endgroup$
    – kvmu
    Dec 23, 2013 at 17:37
  • $\begingroup$ I've added that we can't use calculus for the solution. $\endgroup$
    – user26486
    Dec 23, 2013 at 17:39
  • $\begingroup$ You might get nicer algebra out of writing $a=2+x$, $b=2-x$; then more of the terms will cancel out -- in particular every odd-degree term vanishes. $\endgroup$ Dec 23, 2013 at 18:05
  • $\begingroup$ Related : math.stackexchange.com/questions/487486/… $\endgroup$ Dec 24, 2013 at 13:41

2 Answers 2


You can do this with the quadratic-arithmetic mean: (this is possible, because $a^2\geq 0$.) $$ \sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2\\ a^2+b^2\geq \frac{(a+b)^2}2=8 $$ Now, you only have to proof $\frac 1{a^2}+\frac 1{b^2}\geq \frac 12$. Just as before, we know that $$ \frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2 $$ We know that the minimum of $\frac 1a+\frac 1{4-a}$ is at $a=2$, with outcome $1$. (This can be done by differentiation, or multiply with $a(4-a)$ first.) Now we get $$ \frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2\geq \frac 12 $$

  • 1
    $\begingroup$ $a,b$ aren't positive, though. So we can't use any mean inequalities for them. We could use them for $a^2,b^2$, though. $\endgroup$
    – user26486
    Dec 23, 2013 at 17:43
  • $\begingroup$ @mathh Good point $\endgroup$
    – Ragnar
    Dec 23, 2013 at 17:44
  • $\begingroup$ $a,b$ have to be positive too. We're using the quadratic-arithmetic mean inequality for two variables $a,b$ and all mean inequalities are defined to work only for positive $a$ and $b$. $\endgroup$
    – user26486
    Dec 23, 2013 at 17:48
  • $\begingroup$ @mathh But we can apply them to $|a|$ and $|b|$ and use $|a|\geq a$ and $|b|\geq b$. $\endgroup$
    – Ragnar
    Dec 23, 2013 at 17:49
  • $\begingroup$ Oh. Good point. $\sqrt{\frac{a^2+b^2}2}$ doesn't change its value when the signs of $a$ and $b$ are changed and we know that $\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2$ when $a,b$ are positive and that when $a,b$ change their signs, $\frac{a+b}{2}$ gets a lower value, so it'll still be less or equal to $\sqrt{\frac{a^2+b^2}2}$ $\endgroup$
    – user26486
    Dec 23, 2013 at 17:57

Without loss of generality, we can assume $a\ge b$, so let's write $a=2+x$, $b=2-x$ with $0\le x$. The inequality is clearly satisfied if $a\ge4$, so we need only worry about the range $0\le x\lt2$. Plugging this into the OP's inequality and simplifying like crazy, we find we need only prove

$$f(x)=x^2+{(4+x^2)\over(4-x^2)^2}\ge{1\over4}\quad\text{for }0\le x\lt2$$

Now it does not require calculus to see that the function $f(x)$ is increasing on the interval $0\le x\lt2$: The $x^2$ term is clearly increasing, and so is the quotient term, since the numerator $4+x^2$ is increasing and the denominator $(4-x^2)^2$ is decreasing. Finally, since $f(0)={1\over4}$, we can conclude that the inequality holds.


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