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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..

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

3 Answers 3

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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 $t=2^{-\frac12}\implies a=b=2^{-\frac 34}$.

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    $\begingroup$ (+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 $a=b=1/2^{3/4}$. $\endgroup$
    – Clement C.
    Mar 30, 2021 at 23:19
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    $\begingroup$ 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 $a=b$. $\endgroup$
    – Clement C.
    Mar 30, 2021 at 23:23
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    $\begingroup$ Good point. Amended my answer to reflect that. $\endgroup$ Mar 30, 2021 at 23:35
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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 $2a^2+\frac14a^{-2}$?

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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.

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