We have : \(\sqrt{xy\left(x-y\right)}=x+y\Leftrightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)

\(xy\left(x-y\right)^2=\dfrac{1}{4}.4xy\left[\left(x+y\right)^2-4xy\right]\le\dfrac{\left(x+y\right)^4}{16}\)

so \(\left(x+y\right)^4\ge16\left(x+y\right)^2\)  \(\Leftrightarrow p^4-16p^2\ge0\Leftrightarrow p\ge4\)

Equal sign occurs \(\Leftrightarrow x=2+\sqrt{2};b=2-\sqrt{2}\)

\(P=\left(\dfrac{30}{x}+\dfrac{6x}{5}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)+\dfrac{4}{5}\left(x+y\right)\ge2.\sqrt{\dfrac{30}{x}+\dfrac{6x}{5}}+2.\sqrt{\dfrac{y}{5}+\dfrac{5}{y}}+\dfrac{4}{5}.10\)

\(P=2.6+2.1+8=22\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{30}{x}=\dfrac{6x}{5}\\\dfrac{y}{5}=\dfrac{5}{y}\\x+y=10\end{matrix}\right.=>\left\{{}\begin{matrix}x^2=25\\y^2=25\\x+y=10\end{matrix}\right.\)

\(=>x=y=5\)

\(Pmin=22< =>x=y=5\)

\(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\left(x+y\right)\left(\dfrac{5}{x}+\dfrac{5}{y}\right)+\left(\dfrac{25}{x}+x\right)\)

\(\left\{{}\begin{matrix}x>0=>x=\left(\sqrt{x}\right)^2\\y>0=>y=\left(\sqrt{y}\right)^2\end{matrix}\right.\)

\(P1=x+y\ge10\)

\(P2=\dfrac{5}{x}+\dfrac{5}{y}=5.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge5.\dfrac{4}{x+y}=\dfrac{5.4}{10}=2\) khi \(x=y\)

\(P3=\dfrac{25}{x}+x\ge2.\sqrt{\dfrac{25}{x}.x}=2.5=10\) khi \(x=5\)

\(P=\sum P\ge10+2+10=22\) khi \(\left(x;y\right)=\left(5;5\right)\)

Study well

Reference Lê Anh Duy

5 days

Pls vote for me

\(x^2+5y^2-4xy-x+2y-6=0\)

\(\Leftrightarrow x^2-4xy+4y^2-\left(x-2y\right)-6=-y^2\)

\(\Leftrightarrow\left(x-2y\right)^2-\left(x-2y\right)-6=-y^2\le0\)

\(\Leftrightarrow\left(x-2y+2\right)\left(x-2y-3\right)\le0\)

\(\Leftrightarrow-2\le x-2y\le3\)

Mininum divisors? That is the number 1, obviously

Or did you mean the minimum NUMBER of divisors?
If so than it's 9, how you ask?
To create less permutations with the divisors, the factors should overlap the most, hence, the 6 divisor number is a⁵ and the 4 divisor number is a³, so the product would be a⁸, having 9 divisors.

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

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 \(a^2+b^2\ge5\)

=> \(9-\left(a^2+b^2\right)\le4\)

=> \(2ab\le4\)

=> \(ab\le2\)

<=> \(a^2b^2\ge4\)

=> \(4a^2b^2\ge16\)

(*) 1+N=.....=x2/(x-2)2+4

We have x2 ≥

 0 , (x-2)2+4 ≥

 4 > 0 

So 1+N ≥

 0 => N ≥

 -1 ;equality : x=0 

(*)1-N=....=(x-4)2/(x-2)2+4 

....... -> N ≤

 1 , equality : x=4

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

trình bày = tiếng việt đc ko tiếng anh ngại viết lắm