MHB Find the Minimum Value of a Complex Equation

[Albert1]

$a>1,b>1$

find the minimum value of

$\dfrac {a^2}{b-1}+\dfrac {b^2}{a-1}$

 

[I like Serena]

$a>1,b>1$

find the minimum value of

$\dfrac {a^2}{b-1}+\dfrac {b^2}{a-1}$

Spoiler

Both partial derivatives have to be zero.

From those equations it follows that:
\begin{cases}2a(a-1)^2&=b^2(b-1) \\ a^2(a-1)&=2b(b-1)^2\end{cases}
Multiplying those equations gives us:
$$2a^3(a-1)^3=2b^3(b-1)^3 \Rightarrow a(a-1) = b(b-1)$$
Substitute back into the equations to find:
$$\begin{cases}2a(a-1)^2&=ab(a-1) \\ ab(b-1)&=2b(b-1)^2\end{cases}
\Rightarrow \begin{cases}2(a-1)&= b\\ a&=2(b-1)\end{cases}
\Rightarrow a = b = 2$$
The corresponding minimum value is $8$.

 

[Albert1]

Spoiler

Both partial derivatives have to be zero.

From those equations it follows that:
\begin{cases}2a(a-1)^2&=b^2(b-1) \\ a^2(a-1)&=2b(b-1)^2\end{cases}
Multiplying those equations gives us:
$$2a^3(a-1)^3=2b^3(b-1)^3 \Rightarrow a(a-1) = b(b-1)$$
Substitute back into the equations to find:
$$\begin{cases}2a(a-1)^2&=ab(a-1) \\ ab(b-1)&=2b(b-1)^2\end{cases}
\Rightarrow \begin{cases}2(a-1)&= b\\ a&=2(b-1)\end{cases}
\Rightarrow a = b = 2$$
The corresponding minimum value is $8$.

very good I like Serena !
another solution:

Spoiler

using $AP\geq GP$
$\dfrac {a^2}{b-1}+\dfrac {b^2}{a-1}\geq 2\sqrt {\dfrac {a^2}{a-1}}\times \sqrt{\dfrac {b^2}{b-1}}--(1)$
minimum occurs when $a=b$
let:$\sqrt {\dfrac {a^2}{a-1}}=\sqrt{\dfrac {b^2}{b-1}}=k>0$
we have :$a^2=k^2(a-1)=b^2=k^2(b-1)$
$\rightarrow a^2 - k^2a+k^2=b^2-k^2b+k^2=0--(2)$
for $a,b\in R$
we must have:$k^4-4k^2\geq 0$
or $k^2(k^2-4)\geq 0,\rightarrow k\geq 2$
this implies : $ (1)\geq 2\times 2\times 2=8$
$\therefore $ the minimum of $(1)=8$