# Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$

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.

• A very nice argument based on convexity of $f(x)=\left(x+\frac1x\right)^2$ was given in this comment. Commented May 31, 2016 at 8:08
• Have you tried substituting $q=1-p$? Commented Sep 13, 2019 at 18:22
• Is the answer 12.5 Commented Sep 13, 2019 at 18:34
• 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. Commented Sep 13, 2019 at 18:36
• 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.
– robjohn
Commented Sep 13, 2019 at 18:40

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.

• Although evident, it should be instructed to say that one needs the inequalities for $x,y \in \mathbb{R}$, $2xy \leq x^2 + y^2$ and $4xy \leq (x+y)^2$. Commented Feb 3, 2020 at 23:10

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
• from where did you got the "-15a + 47/4" ? Commented Sep 8, 2013 at 17:27
• From tangent line at 1/2 Commented Sep 8, 2013 at 17:29

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 $S=1,n=2$

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.

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

• 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. Commented May 31, 2016 at 8:32

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 $$s=p(1-p)\leq\frac{1}{4}$$.

• 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? Commented Sep 13, 2019 at 18:34
• 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 Commented Sep 13, 2019 at 18:36
• 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. Commented Sep 13, 2019 at 18:43

Given $$p+q=1$$

Observe that $$\dfrac{p+q}{2} \geq \sqrt{pq}$$ Therefore :- $$\dfrac1{4} \geq {pq}$$ Also, $$p^2+q^2 \geq 2pq$$ similarily, $$\big(\dfrac1p\big)^2+\big(\dfrac1q\big)^2 \geq \dfrac2{pq}$$

Adding both the inequalities We can see $$\big(p+\dfrac1p\big)^2+\big(q+\dfrac1q\big)^2 \geq \dfrac2{pq}+2{pq}+4$$

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.

• Can you explain how, 1/p2+1/q2≥2/pq? Commented Sep 14, 2019 at 13:49
• Divide the first inequality by $pq$ since $pq$ is positive the inequality wont reverse and then whole square both side and the result follows Commented Sep 14, 2019 at 14:44
• Please upvote my answer if you like it .Thanks Commented Sep 14, 2019 at 14:45
• Or you just divide $p^2+q^2 \geq 2pq$ it by $(pq)^2$ Commented Sep 14, 2019 at 14:48
• Oh. Got it. Thanks! I did upvote, but since my reputation is under 15 (new here) so it wouldn't show. Thanks once again! Commented Sep 14, 2019 at 15:38

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.

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 $$1-4pq\geq 0$$

• 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. Commented Sep 13, 2019 at 18:53

If you know the Cauchy-Schwarz inequality, you can use it. Indeed, $$LHS \ge \dfrac{1}{2}\left(p+\dfrac{1}{p}+q+\dfrac{1}{q}\right)^2= \dfrac{1}{2}\left(1+\dfrac{1^2}{p}+\dfrac{1^2}{q}\right)^2\ge\dfrac{1}{2}\left(1+\dfrac{(1+1)^2}{p+q}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2= \dfrac{25}{2} = RHS.$$

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.

• It should probably be explained why $p=q=\frac{1}{2}$. Not obvious a priori (we could have $p$ and $q$ solution of $p+\frac{1}{p}=1-\frac{1}{q^2}$ and $q+\frac{1}{q}=1-\frac{1}{p^2}$, with $p+q=1$) Commented Mar 7, 2023 at 7:30