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Suppose that $a_1,a_2 \in \mathbb{R}$

When does $$2\sqrt{a_1a_2}= a_1+a_2$$

I can't see any solution other than $a_2 = a_1$.

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    Note that this is an instance of the arithmetic-mean geometric-mean inequality but with equality, which is well-known to hold iff the inputs are all equal. (This is a comment rather than an answer.) That is, more generally $n (a_1 \dots a_n)^{1/n} = a_1+\dots+a_n$ holds iff all the $a_i$ are equal. – 2017-02-18
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    @PatrickStevens equal... and non-negative :) – 2017-02-18
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    Note: You can also write it like this ... $2\sqrt{a_1a_2} = \sqrt{a_1}^2 + \sqrt{a_2}^2$ – 2017-02-21

4 Answers 4

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Solving an equation involving square roots is often tricky. Function ‘$\sqrt{\quad}$’ is defined for non-negative reals and it evaluates to non-negative reals, so you have $\sqrt{\quad}: [0,+\infty) \to [0,+\infty)$. When you need to solve an equation like $$2\sqrt{a_1a_2}=a_1+a_2$$ I'd suggest immediatly adding ‘where $a_1a_2 \geq 0$’. From a logical point of view it's absolutely meaningless (let alone you'll often find a wrong solution) trying to solve such an equation where the product $a_1a_2$ is potentially negative. Having in mind all these my derivation would be as follows.

For $a_1, a_2$ such that $a_1a_2 \geq 0$ we have: $$2\sqrt{a_1a_2}=a_1+a_2 \iff$$ $$\{ 4a_1a_2 = (a_1+a_2)^2 \quad \& \quad a_1+a_2 \geq 0 \} \iff$$ $$\{ 4a_1a_2 = a_1^2 + a_2^2 + 2a_1a_2 \quad \& \quad a_1+a_2 \geq 0 \} \iff$$ $$\{ a_1^2 + a_2^2 - 2a_1a_2 = 0 \quad \& \quad a_1+a_2 \geq 0 \} \iff$$ $$\{ (a_1 - a_2)^2 = 0 \quad \& \quad a_1+a_2 \geq 0 \} \iff$$ $$\{ a_1 = a_2 \quad \& \quad a_1+a_2 \geq 0 \} \iff$$ $$a_1 = a_2 \geq 0.$$ (Last step: check that if $a_1 = a_2 \geq 0$, then $a_1a_2 \geq 0$, so we do not reject any solutions.)

Note that the condition ‘$a_1+a_2 \geq 0$’ was necessary for the first equivalence to hold. This is a technique I recommend in order to avoid faulty reasoning when solving such equations while being logically precise and correct!

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If $2\sqrt{a_1a_2}=a_1+a_2$ for some $a_1,a_2\in\mathbb{R}$, then $4a_1a_2=(a_1+a_2)^2$ that is $a_1^2-2a_1a_2+a_2^2=0=(a_1-a_2)^2$, so necessarily $a_1=a_2$. The converse is true if $a_1,a_2\in\mathbb{R}^+$.

So the equality holds iff $a_1=a_2$ and $a_1,a_2\in\mathbb{R}^+$. There is no other solutions.

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    The converse is NOT true, unless you assume that $a_1\ge 0$. Take $a_1=a_2=-1$ – 2017-02-18
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    @MercyKing Thank you, edited. – 2017-02-18
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Indeed you have (squaring both members):

$4a_1a_2=a_1^2+a_2^2+2a_1a_2$

and hence:

$(a_1-a_2)^2=0$

Which gives $a_1=a_2$

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    I'm sad because this didn't hold then the equation which predicts probability of life on other planets would lead to a non-zero solution. – 2017-02-18
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If $2\sqrt{a_1a_2}=a_1+a_2$ so $a_1+a_2\geq0$, which says that we can use squaring.

Hence, for $a_1+a_2\geq0$ we obtain $$2\sqrt{a_1a_2}=a_1+a_2\Leftrightarrow4a_1a_2=(a_1+a_2)^2,$$ which gives $a_1=a_2$ and since $a_1+a_2\geq0$, we get the answer:

$\{(t,t)\}$, where $t\geq0$.