# minimum value of $a^2 + b^2 + \tfrac{1}{(a + b)^2}$

Let $$a$$ and $$b$$ be positive real numbers. Find the minimum value of $$a^2 + b^2 + \frac{1}{(a + b)^2}.$$

So I really have no idea how to start with this one. I've tried using AM-GM to try and cancel out the $$(a+b)^2$$: $$\frac{(a+b)^2-2ab+\tfrac{1}{(a+b)^2}}{3} \geq \sqrt[3]{-2ab}$$ $$a^2+b^2+\frac{1}{(a+b)^2} \geq -3\sqrt[3]{2ab}$$ However this doesn't seem much better..

• Unless I’m missing something, you can trivially bound the expression from below by $0$, which is better than the bound you have given. Mar 30, 2021 at 23:11
• Plugging in $(1,1)$ gives $2.25$, and plugging in $(.5, .5)$ gives $1.5$, and it is bounded below by $0$ Mar 30, 2021 at 23:14

Well, $$(a+b)^2=a^2+b^2+\color{green}{2ab}\leq 2(a^2+b^2)$$ (by AM-GM) which in turn gives $$\frac{1}{(a+b)^2}\geq\frac{1}{2(a^2+b^2)}$$

which gives $$a^2+b^2+\frac{1}{(a+b)^2}\geq a^2+b^2+\frac{1}{2(a^2+b^2)}$$

So now the question is reduced to (setting $$t=a^2+b^2>0$$) : $$\min_{t\in\Bbb R^+}(t+\frac{1}{2t})=\sqrt2$$

Explaining achievability per Clement's comment.

Equality for $$a^2+b^2\geq2ab$$ as seen in the first line (and there on) happens $$\iff a=b$$. This is not disrupted later (we did not, for example, divide through by $$a-b$$) so the minimum is attainable. Indeed, the $$t$$ which minimises the given expression is $$t=2^{-\frac12}\implies a=b=2^{-\frac 34}$$.

• (+1) To be complete, though, the answer should show that this is achievable (i.e., the inequalities are tight at the minimum, nothing was lost in the minoration), for $a=b=1/2^{3/4}$. Mar 30, 2021 at 23:19
• In general, it is worth noting that since the expression is symmetric in $a,b$, chances are it'll be either minimized or maximized at some point on the line $a=b$. Mar 30, 2021 at 23:23
• Good point. Amended my answer to reflect that. Mar 30, 2021 at 23:35

Actually, you are applying AM-GM with negative terms, which just isn't true. ($$2,-1,-1$$ for an example)

Suppose that we have fixed $$a+b$$. Can you show that the minimum is achieved when $$a=b$$?

Can you now minimise $$2a^2+\frac14a^{-2}$$?

Consider $$x^2 + b^2 + \frac{1}{(x + b)^2}$$

Take the derivative wrt. $$x$$, we get

$$2x - \frac{2x}{(x + b)^4}$$

This is zero at $$x=0$$. And it could be one of the minimums... Hence set $$x=0$$, and further set $$b=y$$

$$y^2 + \frac{1}{y^2}$$

Set again the derivative to zero $$2y-\frac{2y}{y^4}=0\Rightarrow y^3=1$$

Hence $$y=1$$

We get $$2$$ as a possible infimum. If you check the other zeroes it becomes evident that it is the minimum. In particlar, if $$x$$ or $$y$$ belongs in $$(0,1]$$, the minimum is attained.