Question

[Minh Hoàng Vũ]

cho 0≤a,b,c≤3 và a+b+c=4. tìm min,maxP= a^4+b^4+c^4

Answer 1

[Lê Song Phương]

Có \(a^4+b^4+c^4=\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3}\) 

\(\ge\dfrac{\left(\dfrac{\left(a+b+c\right)^2}{3}\right)^2}{3}\)  (áp dụng 2 lần BĐT \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\))

\(=\dfrac{\left(\dfrac{4^2}{3}\right)^2}{3}=\dfrac{256}{27}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\dfrac{4}{3}\)

Vậy \(minP=\dfrac{256}{27}\) khi \(a=b=c=\dfrac{4}{3}\)

Answer 2

[Nguyễn Việt Lâm]

Min P dễ em có thể tự tìm đơn giản bằng AM-GM

\(P=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ca\right)^2+4abc\left(a+b+c\right)\)

\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ca\right)^2+16abc\)

Do \(0\le a;b;c\le3\Rightarrow\left(3-a\right)\left(3-b\right)\left(3-c\right)\ge0\)

\(\Rightarrow3\left(ab+bc+ca\right)-9\left(a+b+c\right)+27-abc\ge0\)

\(\Rightarrow ab+bc+ca\ge\dfrac{abc+9}{3}\)

\(a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=16-2\left(ab+bc+ca\right)\)

\(\le16-\dfrac{2}{3}\left(abc+9\right)\)

Do đó:

\(P\le\left[16-\dfrac{2}{3}\left(abc+9\right)\right]^2-2\left(\dfrac{abc+9}{3}\right)^2+16abc\)

Đặt \(abc=x\Rightarrow0\le x\le\dfrac{64}{27}\)

\(P\le\left[16-\dfrac{2}{3}\left(x+9\right)\right]^2-2\left(\dfrac{x+9}{3}\right)^2+16x\)

\(P\le\dfrac{2}{9}\left(x^2-6x+369\right)\)

\(P\le\dfrac{2}{9}x\left(x-6\right)+82\)

Do \(0\le x\le\dfrac{64}{27}\Rightarrow x-6< 0\Rightarrow\dfrac{2}{9}x\left(x-6\right)\le0\)

\(\Rightarrow P\le82\)

Dấu "=" xảy ra khi \(x=0\) hay \(\left(a;b;c\right)=\left(0;1;3\right)\) và các hoán vị