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If $a, b$ are positive real numbers and $a+b = 1$, prove that :

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

I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.

Thank you.

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    $\begingroup$ A very nice argument based on convexity of $f(x)=\left(x+\frac1x\right)^2$ was given in this comment. $\endgroup$ – Martin Sleziak May 31 '16 at 8:08
  • $\begingroup$ Have you tried substituting $q=1-p$? $\endgroup$ – saulspatz Sep 13 at 18:22
  • $\begingroup$ Is the answer 12.5 $\endgroup$ – Akshaj Bansal Sep 13 at 18:34
  • $\begingroup$ I did, and I differentiated it once and equated to zero, in the hopes of obtaining an extreme, but it became quite complicated and confused me. $\endgroup$ – shreyassps Sep 13 at 18:36
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    $\begingroup$ Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. $\endgroup$ – robjohn Sep 13 at 18:40

10 Answers 10

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First Method.

$a^{2}+\dfrac{1}{a^{2}}\geq -15a+\dfrac{47}{4}$ $~$ $\Longleftrightarrow$ $~$ $(2a-1)^{2}(a^{2}+16a+4)\geq 0$ : evident
$\therefore$ $\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+a^{2}+\dfrac{1}{a^{2}}+b^{2}+\dfrac{1}{b^{2}}\geq 4-15a+\dfrac{47}{4}-15b+\dfrac{47}{4}=\dfrac{25}{2}$

Second Method.

$\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+\underline{a^{2}+b^{2}}+\underline{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}}\geq 4+\underline{\dfrac{1}{2}}+\underline{8}=\dfrac{25}{2}$

  1. $(1^{2}+1^{2})(a^{2}+b^{2})\geq (a+b)^{2}$ : Cauchy-Schwarz
  2. $(a+b)(a+b)\left(\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}\right)\geq (1+1)^{3}$ : Holder
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  • $\begingroup$ from where did you got the "-15a + 47/4" ? $\endgroup$ – user93765 Sep 8 '13 at 17:27
  • $\begingroup$ From tangent line at 1/2 $\endgroup$ – chloe_shi Sep 8 '13 at 17:29
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First $$(a+1/a)^2 + (b+1/b)^2 \geq\frac{1}{2} (a+b+1/a+1/b)^2=\frac{1}{2}(1+1/(ab))^2.$$

Then note that $$ab\le(a+b)^2/4=1/4.$$

Take it into the first one, you may get your inequality.

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Generalization:

If $\sum_{1\le r\le n}a_r=S$ where $a_i$s are positive real numbers

$$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2=\sum_{1\le r\le n}a_r^2+\sum_{1\le r\le n}a_r^{-2}+2n$$

We know $$\frac{\sum_{1\le r\le n}a_r^m}n> \text{ or } <\left(\frac{\sum_{1\le r\le n}a_r}n\right)^m$$ according as $m$ lies or does not lie in $(0,1)$

Putting $m=2,$ $$\frac{\sum_{1\le r\le n}a_r^2}n>\left(\frac{\sum_{1\le r\le n}a_r}n\right)^2=\left(\frac Sn\right)^2=\frac{S^2}{n^2}$$

Putting $m=-2,$ $$\frac{\sum_{1\le r\le n}a_r^{-2}}n>\left(\frac{\sum_{1\le r\le n}a_r}n\right)^{-2}=\left(\frac Sn\right)^{-2}=\frac{n^2}{S^2}$$

On simplification, $$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2\ge \frac{(n^2+S^2)^2}{S^2n}$$

Here $S=1,n=2$

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We can use the inequalities between quadratic mean (a.k.a. square root mean), arithmetic mean, geometric mean and harmonic mean $$\sqrt{\frac{x^2+y^2}2} \ge \frac{x+y}2 \ge \sqrt{xy} \ge \frac2{\frac1x+\frac1y}.$$ These inequalities are true for any $x,y>0$. The equality holds if and only if $x=y$. They can be generalized for more than two variables (also proofs for two variables are easier). If you are not familiar with these inequalities, you can find a lot of material on them, just try to search for some reasonable queries like this one or this one.

You want to minimize $f(a,b)=\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2$. Notice that $$f(a,b)=\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2 = a^2+b^2 + \frac1{a^2}+\frac1{b^2} +4.$$ So minimizing $g(a,b)=a^2+b^2 + \frac1{a^2}+\frac1{b^2}$ is an equivalent problem.

Using QM-AM inequality we get $$\frac{a^2+b^2}2 \ge \left(\frac{a+b}2\right)^2 = \frac14$$ which implies $$a^2+b^2 \ge \frac12.\tag{1}$$

If we combine HM-GM and AM-GM we get $$\frac2{\frac1{a^2}+\frac1{b^2}} \le \sqrt{a^2b^2} = ab \le \left(\frac{a+b}2\right)^2 = \frac14,$$ which is equivalent to $$\frac1{a^2}+\frac1{b^2} \ge 8.\tag{2}$$

By adding the inequalities $(1)$ and $(2)$ we get $$g(a,b) = a^2+b^2 + \frac1{a^2}+\frac1{b^2} \ge 8+\frac12$$ and $$f(a,b)=g(a,b)+4 \ge 12+\frac12 = \frac{25}2.$$


The inequalities $(1)$ and $(2)$ can be also interpreted geometrically.

Minimizing $a^2+b^2$ for $a+b=1$ is simply finding the point on the line $a+b=1$ which is closest to the origin. If you draw the picture, you immediately see that it is the point given by $a=b=\frac12$.

We also want some geometrical insight into minimizing $\frac1{a^2}+\frac1{b^2}$ for $a+b=1$. Let us transform this problem a bit.

If we denote $x=\frac1a$, then $y=\frac1b=\frac1{1-a}=\frac1{1-\frac1x}=\frac{x}{x-1}$. It is not difficult to see that this is a hyperbole with the asymptotes $x=1$ and $y=1$. And we want to minimize $\frac1{a^2}+\frac1{b^2}=x^2+y^2$, which means that we want to find the closest point to the origin on this hyperbole. (To be more precise, only in the part of the hyperbole in the first quadrant, since $x,y>0$.) Here is a plot from WolframAlpha.

Hyperbole - plot from WA

Again we see from the picture that the closest point will be the one with $x=y$. (Which gives us $x=y=2$ and $a=b=\frac12$.)

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    $\begingroup$ Originally I posted this as an answer to this question. But since the emphasis of that question seems to be evaluation of the OPs attempt, this is probably a better place where to post this. $\endgroup$ – Martin Sleziak May 31 '16 at 8:32
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Let $f(x)=\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$. Note that $f(x)$ is convex on the $(0;+\infty)$ because its second derivative is equal to $$ f''(x)=2+\frac{6}{x^4}>0,~\text{for all}~x\in(0;+\infty). $$ Hence, by Jensen's inequality ($p+q=1$) $$ f(p)+f(q)\geqslant 2f\left(\frac{p+q}{2}\right)=2f\left(\frac{1}{2}\right)=\frac{25}{2}. $$ Hence, $$ \left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2\geq \frac{25}{2}. $$ Thus, the correct answer is C.

Here is an elementary approach. We will prove that for $p\in(0;1)$ the following inequality holds $$ \left(p+\frac{1}{p}\right)^2+\left(1-p+\frac{1}{1-p}\right)^2\geq \frac{25}{2}, $$ or $$ p^2+(1-p)^2+\frac{1}{p^2}+\frac{1}{(1-p)^2}\geq \frac{17}{2}, $$ or $$ (p^2+(1-p)^2)\left(\frac{p^2(1-p)^2+1}{p^2(1-p)^2}\right)\geq \frac{17}{2}. $$ Now, note that $$ p^2+(1-p)^2=(p+(1-p))^2-2p(1-p)=1-2p(1-p). $$ Denote $s=p(1-p)$. Thus, we need to prove that if $p\in(0;1)$ then $$ (1-2s)\frac{s^2+1}{s^2}\geq\frac{17}{2}, $$ or $$ 2(1-2s)(s^2+1)\geq 17s^2. $$ The last inequality is equivalent to $$ 4s^3+15s^2+4s-1\leq 0. $$ However, $4s^3+15s^2+4s-2=(4s-1)(s^2+4s+2)\leq 0$ because $s=p(1-p)\leq\frac{1}{4}$.

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  • $\begingroup$ Hi. Thanks for the swift reply. I have no clue about Jensen's inequality, so could you explain about it? And is there a way without it? $\endgroup$ – shreyassps Sep 13 at 18:34
  • $\begingroup$ Jensen's inequality is a general statement for convex functions. You can find it in any course of differential calculus. More information you can find here en.wikipedia.org/wiki/Jensen%27s_inequality $\endgroup$ – richrow Sep 13 at 18:36
  • $\begingroup$ Is there a way to solve it without this inequality? I'm trying to find the minima by differentiating, but I can't seem to get it right. $\endgroup$ – shreyassps Sep 13 at 18:43
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Here is how to solve it as illustrated with Matlab so simplify the steps. The first step is to substitute q=1-p into the second term in your sum. Then this is differentiated w.r.t. p, set equal to zero, and solved for the value of p which gives the minimum. This value is then used to find the sum. So p = 1/2 = q, and the minimum is 12.5.

enter image description here

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Given $p+q=1$

Observe that $\frac{p+q}{2} \geq \sqrt{pq}$ Therefore :- $\frac1{4} \geq {pq}$ Also, $p^2+q^2 \geq 2pq$ similarily, $(\frac1p)^2+(\frac1q)^2 \geq \frac2{pq}$

Adding both the inequalities We can see $\big(p+\frac1p\big)^2+\big(q+\frac1q\big)^2 \geq \frac2{pq}+2{pq}+4$

You can see for getting minimum value the left hand side should be as minimum as possible which holds when $pq$ is maximum Therefore substituting the maximum value of $pq$ that is $\frac14$ we get the answer:$12.5$

Thanks.

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  • $\begingroup$ Can you explain how, 1/p2+1/q2≥2/pq? $\endgroup$ – shreyassps Sep 14 at 13:49
  • $\begingroup$ Divide the first inequality by $pq$ since $pq$ is positive the inequality wont reverse and then whole square both side and the result follows $\endgroup$ – Akshaj Bansal Sep 14 at 14:44
  • $\begingroup$ Please upvote my answer if you like it .Thanks $\endgroup$ – Akshaj Bansal Sep 14 at 14:45
  • $\begingroup$ Or you just divide $p^2+q^2 \geq 2pq$ it by $(pq)^2$ $\endgroup$ – Akshaj Bansal Sep 14 at 14:48
  • $\begingroup$ Oh. Got it. Thanks! I did upvote, but since my reputation is under 15 (new here) so it wouldn't show. Thanks once again! $\endgroup$ – shreyassps Sep 14 at 15:38
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Hint: It is $$\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2\geq \frac{25}{2}$$ This is equivalent to $$(p^2+q^2)(1+\frac{1}{p^2q^2})\geq \frac{17}{2}$$ With $$p^2+q^2=1-2pq$$ we get $$1-2pq+\frac{1}{p^2q^2}-\frac{2}{pq}\geq \frac{17}{2}$$ This is equivalent to $$(1-4pq)(p^2q^2+4pq+2)\geq 0$$ And by AM-GM we get $$\frac{p+q}{2}\geq \sqrt{pq}$$ or $$1-4pq\geq 0$$

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  • $\begingroup$ Hi. Thank you sir for the reply. But this seems like you started with the equation being greater than or equal to 12.5, Could you please elaborate on this? I'm sorry for a rookie question. $\endgroup$ – shreyassps Sep 13 at 18:53
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If you know the Cauchy-Schwarz inequality, you can use it. Indeed, $LHS \ge \dfrac{1}{2}\left(p+\dfrac{1}{p}+q+\dfrac{1}{q}\right)^2= \dfrac{1}{2}\left(1+\dfrac{1^2}{p}+\dfrac{1^2}{q}\right)^2\ge\dfrac{1}{2}\left(1+\dfrac{(1+1)^2}{p+q}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2= \dfrac{25}{2} = RHS. $

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Let $f(p,q)=\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2$.

Using Lagrange multiplier's method: $$L=\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2+\lambda (1-p-q)\\ \begin{cases}L_p=2\left(p+\frac1p\right)(1-\frac1{p^2})-\lambda=0\\ L_q=2\left(q+\frac1q\right)(1-\frac1{q^2})-\lambda=0\\ L_{\lambda}=1-p-q=0\end{cases} \Rightarrow p=q=\frac12$$ SOC: $$L_{pp}=2\left(1-\frac1{p^2}\right)^2+2\left(p+\frac1p\right)\cdot \frac1{p^3}>0\\ L_{pq}=0\\ \Delta=L_{pp}L_{qq}-L_{pq}^2=L_{pp}^2>0$$ Hence: $f(\frac12,\frac12)=12.5$ is minimum.

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