# If $a+\sqrt{a}=b+\sqrt{b}$ is $a=b$?

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?

• Hint: what is the derivative of $f(x) = x+x^{1/2}$? Aug 21 '20 at 22:42
• @Integrand ohhh Aug 21 '20 at 22:45

If $$a>b$$ then $$\sqrt{a}>\sqrt{b}$$ and $$LHS>RHS$$. If $$a then $$LHS. Thus $$a=b$$.

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 $$a=b$$

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

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 $$f(a) = f(b) \implies a = b$$.