Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$ 
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$

My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$ 
Seems to using Cauchy-Schwartz. Please help.
 A: Using Cauchy-Schwarz Inequality we have that:
$$\left(\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\right)(1 + 1) \ge \left( a + b + \frac 1a + \frac 1b \right)^2 \ge 5^2 = 25$$
The last inequality follows as:
$$\frac 1a + \frac 1b = \frac{b + a}{ab} = \frac 1{ab} \ge \frac{4}{(a+b)^2} = 4$$

Another way to solve the inequality is to expand the squares and use the fact that:
$$a^2 + b^2 \ge \frac{(a+b)^2}{2} = \frac 12 \quad \text{ and } \quad \frac 1{a^2} + \frac 1{b^2} \ge \frac 1{2a^2b^2} \ge 8$$
A: A shortcut that I found (Correct me if I am wrong in spotting):
Using AM-GM Inequality, we get
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}}$$
$$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
$$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{1}$$
Using Cauchy-Schwarz Inequality,
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2$$
$$\ge \bigg(a+\dfrac{1}{a}+b+\dfrac{1}{b}\bigg)^2$$
$$\ge (1+4)^2$$
$$\ge 25$$
$$\ge \frac{25}{2}$$
Hope this helps.
A: Since $f(x)=\left(x+\frac{1}{x}\right)^2$ is a convex function, your inequality follows from Jensen:
$$LS\geq2\left(\frac{a+b}{2}+\frac{1}{\frac{a+b}{2}}\right)^2=\frac{25}{2}$$
A: Ok, so we have that $a+b=1$.
From AM-GM, this immediately tells us that $ab \leq 0.25$. Furthermore, since you said that $\frac{1}{a} + \frac{1}{b} = \frac{1}{ab}$, this shows that $\frac{1}{a} + \frac{1}{b} \geq 4$.
Now, we are given two numbers $a + \frac 1a$ and $b+ \frac 1b$. Their sum is greater than or equal to $5$, since $a+b=1$. Let their sum be $c$.
Then, by AM-GM, $(a+\frac{1}{a})(b+\frac 1b) \leq \frac{c^2}{4}$, whose minimum value is $6.25$, since $c \geq 5$.
Hence, it follows that $(a+\frac 1a)^2 + (b + \frac 1b)^2 = c^2 - 2(a+\frac{1}{a})(b+\frac 1b) \geq 25 - 12.5 \geq 12.5$, since $c \geq 5$.
This proof required nothing more than just ordinary AM-GM. Sometimes it's nice to think simple.
A: I just differentiate. 
$$(ab+\dfrac1{ab})(2-\dfrac1{ab})=(x+\dfrac1x)(2-\dfrac1x)=f(x)$$
and $x\leq\dfrac14$ 
$f'(x)=(1-\dfrac1{x^2})(2-\dfrac1x)+\dfrac1{x^2}(x+\dfrac1x)
     =\dfrac{2x^3-2x+2}{x^3}
     =2*\dfrac{x^3-x+1}{x^3}>0$
$LHD=4-minf(x)=4-f(\dfrac14)=4-(\dfrac14+4)(-2)=\dfrac{25}2$
A: Calculus will also work. Let $f(a)=(a+\frac {1}{a})^2 +(b+\frac {1}{b})^2$ with $b=1-a$ and, WLOG, $a\geq b.$ So $1/2\leq a<1.$
Since $db/da=-1$ we have $$f'(a)=2\left(a+\frac {1}{a}\right) \left(1-\frac {1}{a^2}\right)+2\left(b+\frac {1}{b}\right) \left(-1+\frac {1}{b^2}\right)=$$ $$=2(a-b)+2\left(\frac {1}{b^3}-\frac {1
}{a^3}\right)=2(a-b)+2\frac {(a^3-b^3)}{ab}$$ which is positive when $a>b>0$, that is, when $1/2<a<1$. 
So $f(a)\geq f(1/2)=25/2.$
