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: 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 $S=1,n=2$
A: 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 $a=b=\frac12$.) 
A: 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 $s=p(1-p)\leq\frac{1}{4}$. 
A: 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.
A: 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.
A: 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. 
$\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+\underline{a^{2}+b^{2}}+\underline{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}}\geq 4+\underline{\dfrac{1}{2}}+\underline{8}=\dfrac{25}{2}$    


*

*$(1^{2}+1^{2})(a^{2}+b^{2})\geq (a+b)^{2}$ : Cauchy-Schwarz

*$(a+b)(a+b)\left(\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}\right)\geq (1+1)^{3}$ : Holder
A: 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.

A: 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 $$1-4pq\geq 0$$
A: 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. $
A: 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.
