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If $a+\sqrt{a}=b+\sqrt{b}$, does this automatically mean that $a=b$?

I first tried to square both sides but that seemed to get me nowhere.

$$a-b=\sqrt{b}-\sqrt{a}$$

Can we just conclude that $a$ has to be equal to $b$ to make this expression to be true?

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    $\begingroup$ Hint: what is the derivative of $f(x) = x+x^{1/2}$? $\endgroup$
    – Integrand
    Aug 21 '20 at 22:42
  • $\begingroup$ @Integrand ohhh $\endgroup$
    – John Rawls
    Aug 21 '20 at 22:45
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If $a>b$ then $ \sqrt{a}>\sqrt{b}$ and $LHS>RHS$. If $a<b$ then $LHS<RHS$. Thus $a=b$.

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Let $a\ge 0$ and $ b\ge 0 $ such that $$a+\sqrt{a}=b+\sqrt{b}$$ then

$$a-b=\sqrt{b}-\sqrt{a}$$ $$=(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})$$

and

$$(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}+1)=0$$

but $$\sqrt{a}+\sqrt{b}+1\ne 0$$

thus, necessarily $$\sqrt{a}=\sqrt{b}$$ and $$a=b$$

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$a-b=\sqrt{a}-\sqrt{b} \implies (\sqrt{a}-\sqrt{b})(-1+\sqrt{a}+\sqrt{b})=0\implies \sqrt{a}=\sqrt{b}$ or $1=\sqrt{a}+\sqrt{b}$. So it might not be the case that $a = b$.

I realize my answer above is for a different problem. So back to this one. We have $a - b = \sqrt{b} - \sqrt{a} \implies a - b +\sqrt{a}-\sqrt{b} = 0 \implies (\sqrt{a}-\sqrt{b})(1+\sqrt{a}+\sqrt{b}) = 0 \implies \sqrt{a} = \sqrt{b} \implies a = b$ .

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Let $f(x) = x + \sqrt x$. Then, we have $f'(x) = 1 + 1/(2\sqrt x) > 0 ~ (\forall x>0)$. Therefore, $f(x)$ is strictly increasing and $f(a) = f(b) \implies a = b$.

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