Fundamental Inequalities - 2

Let $f(a,b)=a^2+b^2+\dfrac{1}{(a+b)^2}$

1) Remark that if $ab<0$ we have $|a+b|<|a|+|b|$ , so $f(a,b)>f(|a|,|b|)$ . Moreover $a$ and $b$ play symmetric roles and $f(-a,-b)=f(a,b)$ , therefore we only care about $(a,b)\in\{(x,y)\in\mathbb{R}^2, \ x\geq 0, \ y\geq x\}\setminus \{(0,0)\}$ .

2) $f$ has indeed a minimum: if $a,b \longrightarrow 0 \ \text{or} \ \left( a\longrightarrow +\infty \ \text{or} \ b\longrightarrow +\infty \right) , \ f(a,b)\longrightarrow +\infty$ so we can reduce the problem to a compact. $f$ is continuous on a compact set thus $f$ has a minimum in this set (Bolzano and Weierstrass theorem) and it is a global minimum due to what is said above.

3) To apply principles of multivariable calculus to our case we need to be on an open set (see note). I will take $\{(x,y)\in\mathbb{R}^2, \ y>-x\}$ , our minimum is reached in this set at -let's say- $(u,v)$ . Thus:

$\left(\dfrac{\partial f}{\partial a}(u,v),\dfrac{\partial f}{\partial b}(u,v)\right)=(0,0)$

Rewriting,

$u(u+v)^3=1\\ v(v+u)^3=1$

So, $u=v$ (by dividing the two equations). Replacing in one equation we get $u^4=\frac{1}{8}$ . So we can take indifferently $u=\pm\frac{1}{\sqrt{2\sqrt{2}}}$ .

Thus the answer is $f\left(\dfrac{1}{\sqrt{2\sqrt{2}}},\dfrac{1}{\sqrt{2\sqrt{2}}}\right)=\boxed{\sqrt{2}}\$ .

Note: If we are not on an open set, $f$ (a differentiable function) has an extremum at $x_0$ does not imply $f'(x_0)=0$ !

Example: $\begin{array}{c}f: & [0;1] & \longrightarrow & \mathbb{R} \\ & x & \longmapsto & x \end{array} \$ at $x_0=1$ or $x_0=0$ .