\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=9\\a^4+2a^2b^2+b^4\ge25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2ab=9-\left(a^2+b^2\right)\\a^4+b^4+2a^2b^2\ge25\end{matrix}\right.\)

We have : a² + b² \(\ge\) 5

=> \(-\left(a^2+b^2\right)\le5\)

=> \(9-\left(a^2+b^2\right)\ge9-5=4\)

=> \(2ab\ge4\)

=> \(ab\ge2\)

<=> \(a^2b^2\ge4\)

<=> \(4a^2b^2\ge16\)

Plus \(4a^2b^2\ge16\) into \(a^4+b^4+2a^2b^2\ge25\)

=> \(a^4+b^4+6a^2b^2\ge41\)

=> Min = 41

That is my opinion :v 

{a+b=3a2+b2≥5⇒{a2+2ab+b2=9a4+2a2b2+b4≥25⇒{2ab=9−(a2+b2)a4+b4+2a2b2≥25

We have : a2 + b2 ≥

 5

=> −(a2+b2)≤5

=> 9−(a2+b2)≥9−5=4

=> 2ab≥4

=> ab≥2

<=> a2b2≥4

<=> 4a2b2≥16

Plus 4a2b2≥16

 into a4+b4+2a2b2≥25

=> a4+b4+6a2b2≥41

=> Min = 41

That is my opinion :v

There is something is wrong here 

Change to :

We have a2+b2≥5

=> 9−(a2+b2)≤4

=> 2ab≤4

=> ab≤2

<=> a2b2≥4

=> 4a2b2≥16

There is something is wrong here 

Change to :

We have \(a^2+b^2\ge5\)

=> \(9-\left(a^2+b^2\right)\le4\)

=> \(2ab\le4\)

=> \(ab\le2\)

<=> \(a^2b^2\ge4\)

=> \(4a^2b^2\ge16\)

Kaya Renger,you were wrong :

\(-\left(a^2+b^2\right)\le-5\Rightarrow9-\left(a^2+b^2\right)\le4\)

Based on your solution, we have \(4a^2b^2\le16;a^4+b^4+2a^2b^2\ge25\)

So, we can't find the minimum value of P.

Maybe your way is not right.

No, it's my mistake , not you :V 

Kaya Renger There is no mistake in the thread.

You can find the answer in here : Câu hỏi của Đức Minh - Toán lớp 9 | Học trực tuyến