6
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ The polynomial has real roots $2$. $\endgroup$
    – Don Larynx
    Dec 23 '13 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 '13 at 17:37
  • $\begingroup$ I've added that we can't use calculus for the solution. $\endgroup$
    – user26486
    Dec 23 '13 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 '13 at 18:05
  • $\begingroup$ Related : math.stackexchange.com/questions/487486/… $\endgroup$ Dec 24 '13 at 13:41
8
$\begingroup$

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 $$ \frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2\geq \frac 12 $$

$\endgroup$
10
  • 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 '13 at 17:43
  • $\begingroup$ @mathh Good point $\endgroup$
    – Ragnar
    Dec 23 '13 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 '13 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 '13 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 '13 at 17:57
4
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.