Cho \(0< x< y\le z\le1\) và \(3x+2y+z\le4\). Tìm Max \(S=3x^2+2y^2+z^2\) - Hoc24

Tham khảo

Khai triển Abel ta có:

\(S=\left(z-y\right)z+\left(y-x\right)\left(z+2y\right)+x\left(3x+2y+z\right)\)

\(\le\left(z-y\right).1+\left(y-x\right).3+4x=x+2y+z\)

\(=\left(1-1\right)z+\left(1-\dfrac{1}{3}\right)\left(2y+z\right)+\dfrac{1}{3}\left(3x+2y+z\right)\)

\(\le\dfrac{2}{3}.3+\dfrac{1}{3}.4=\dfrac{10}{3}\)

Dấu = xảy ra khi \(x=\dfrac{1}{3},y=z=1\)

\(S=x\left(3x+2y+z\right)+\left(y-x\right)\left(2y+z\right)+\left(z-y\right).y\)

\(S\le4x+3\left(y-x\right)+z-y=x+2y+z\)

\(S\le\dfrac{1}{3}\left(3x+2y+z\right)+\dfrac{2}{3}\left(2y+z\right)\le\dfrac{1}{3}.4+\dfrac{2}{3}.3=\dfrac{10}{3}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};1;1\right)\)

Lời giải:

Do \(-2\leq x,y\leq 1\) nên:

\(\left\{\begin{matrix} (x+2)(x-1)\leq 0\\ (y+2)(y-1)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2+x-2\leq 0\\ y^2+y-2\leq 0\end{matrix}\right.\)

\(\Rightarrow x^2+x-2+y^2+y-2\leq 0\)

\(\Leftrightarrow x^2+y^2\leq 4-(x+y)=4\)

Ta có đpcm.

we had abc+(4-a)(4-b)(4-c)\(\ge0\). khai triển ta có \(ab+bc+ca\ge8\)( maybe)

\(P=\left(a+b+c\right)^2-\left(ab+bc+ca\right)\le6^2-8=28\)

Dấu = xảy ra (a,b,c)~(0;2;4) và các hoán vị

Đây nhé @Liana

\(2a^2+\left(2-\sqrt{3}\right)b^2+2a^2+\left(2-\sqrt{3}\right)c^2+\left(\sqrt{3}-1\right)b^2+\left(\sqrt{3}-1\right)c^2\)

\(\ge2\sqrt{4-2\sqrt{3}}ab+2\sqrt{4-2\sqrt{3}}ac+2\left(\sqrt{3}-1\right)bc\)

\(\Leftrightarrow4a^2+b^2+c^2\ge2\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow4\ge2\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow1+\sqrt{3}\ge ab+bc+ca\)

\(\Rightarrow dpcm\)

b)Từ \(a+b+c=6\Rightarrow\left(a+b+c\right)^2=36\)

\(\Rightarrow36=a^2+b^2+c^2+2\left(ab+bc+ca\right)=P+ab+bc+ca\)

\(\Rightarrow P=36-ab-bc-ca\). Cần tìm \(GTNN\) của \(ab+bc+ca\)

Không mất tính tổng quát giả sử \(a=max\left\{a,b,c\right\}\)

\(\Rightarrow a+b+c=6\le3a\Rightarrow2\le a\le4\). Lại có:

\(ab+bc+ca\ge ab+ac=a\left(b+c\right)=a\left(6-a\right)\ge8\)

Suy ra GTNN của \(ab+bc+ca=8\Leftrightarrow\hept{\begin{cases}a=4\\b=2\\c=0\end{cases}}\)

Vậy GTLN_P là \(36-8=28\) khi \(\hept{\begin{cases}a=4\\b=2\\c=0\end{cases}}\)