Wonderful property usage

$a + b = 1$ & $a + b \ge 2\sqrt{ab} \Rightarrow \frac{1}{4} \ge ab \Rightarrow \frac{1}{ab} \ge 4.$

By Cauchy-Schwarz, we get:

$\large (x^{2} + y^{2})(1^{2} + 1^{2}) \ge {(x+y)^{2}} \Rightarrow x^{2} + y^{2} \ge \frac{(x + y)^{2}}{2}$ . $\large (a + \frac{1}{a})^{2} +( b + \frac{1}{b})^{2} \ge \frac{1}{2}[a+b+\frac{1}{a} + \frac{1}{b}]^{2} = \frac{1}{2}(1 + \frac{1}{ab}) \ge \frac{1}{2}(5)^{2} \ge \boxed{\frac{25}{2}}$

We indeed see this minimum is attained when $a = b = \frac{1}{2}$

a + b =1

therefore a right angled triangle will be formed

solution

now let lenth(perpendicular + base) = x

thus if perpendicular is y than base is x -y

thus area of the triangle = $\frac{1}{2}(x)(x - y) = \frac{1}{2}(x^{2} - x y)$

now $f(x) = \frac{1}{2}(x^{2} - x y)$

then $f'(x) = 2x - y$ f'(x) = 0 then $x = \frac{y}{2}$

now $f"(x) = 2$ therefore it is the point of minima thus a = b = 1/2

now keeping the value we get 12

My attempt at a solution without words:

$f(x)=(a+\frac{1}{a} )^{2}+(b+\frac{1}{b} )^{2}$

$=a^{2}+\frac{1}{a^{2}} +2+b^{2}+\frac{1}{b^{2}} +2$

$f^{'}(x)=2a-\frac{2}{a^{3}} +2b-\frac{2}{b^{3}}$

$2a^{4}b^{3}-2b^{3}+2b^{4}a^{3}-2a^{3}=0$

$2(a^{4}b^{3}-b^{3}+b^{4}a^{3}-a^{3})=0$

$\text{(differentiating w.r.t x)}12a^{3}b^{2}-3b^{2}+12a^{2}b^{3}-3a^{2}=0$

$\text{(differentiating w.r.t x)}72a^{2}b-6b+72ab^{2}-6a=0$

$6(12a^{2}b-b+12ab^{2}-a)=0$

$12a^{2}(1-a)-(1-a)+72a(1-a)^{2}-a=0$

$12a^{2}-12a^{3}-1+a+12(1-2a+a^{2})a-a=0$

$12a^{2}-12a^{3}-1+a+12a-24a^{2}+12a^{3}-a=0$

$-12a^{2}+12a-1$

$\text{(differentiating w.r.t x)}-24a+12=0$

$a=\boxed{\frac{1}{2}} b=\boxed{\frac{1}{2}}$

$(0.5+\frac{1}{0.5} )^{2}+(0.5+\frac{1}{0.5} )^{2}$

$2.5^{2}+ 2.5^{2}$

$12.5$

$\left\lfloor 12.5\right\rfloor = \boxed{12}$

To all those who did not like the 3 words I used I am really sorry but they were very important. You may ignore them.

Let $f(x)=(x+\frac{1}{x})^2$ . Since $f(x)$ is a convex function, by Jensen's Inequality, we have $f(\frac{a+b}{2}) \leq \frac{f(a)+f(b)}{2}$ so, we have $f(a)+f(b) \geq 12.5$ and hence $[12.5]=\boxed{12}$

Let √a=cos∆ & √b=sin∆ substitute mimnima occurs at ∆=π/4 .....As simple as that :-)

put a=b=1/2 ,,,,,,,and get ans...........=12.5