What Are the Positive Real Solutions for a and b in this Equation?

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SUMMARY

The equation $a+b+\dfrac{1}{a}+\dfrac{1}{b}+4=2(\sqrt{2a+1}+\sqrt{2b+1})$ has been analyzed, revealing that the positive real solutions for $a$ and $b$ are both equal to 1. Substituting $a = 1$ and $b = 1$ satisfies the equation, confirming that $(1, 1)$ is the only solution. This conclusion is derived from simplifying the equation and applying algebraic manipulation techniques.

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Determine all positive real $a$ and $b$ satisfying the equation $a+b+\dfrac{1}{a}+\dfrac{1}{b}+4=2(\sqrt{2a+1}+\sqrt{2b+1})$.
 
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Solution of other:

Notice that $a+\dfrac{1}{a}+2-2\sqrt{2a+1}=\dfrac{a^2+2a+1-2a\sqrt{2a+1}}{a}=\dfrac{(a-\sqrt{2a+1})^2}{a}$ .

Hence the original equation can be rewritten as

$\dfrac{(a-\sqrt{2a+1})^2}{a}+\dfrac{(b-\sqrt{2b+1})^2}{b}=0$.

For $a,\,b>0$, this gives $a-\sqrt{2a+1}=0$ and $b-\sqrt{2b+1}=0$. It follows that the only solution is $a=b=1+\sqrt{2}$.
 

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