áp dụng bđt bunhia cốp xki ta có cặp số \(\left(a,2b,c\right)\left(1,\sqrt{2},1\right)\)

\(\left(a^2+2b^2+c^2\right)\left(1+\sqrt{2}+1\right)>=\left(a+b+c\right)^2\)

\(a^2+2b^2+c^2>=\frac{0^2}{2+\sqrt{2}}=0\)

dấu "=" xảy ra khi và chỉ khi \(\frac{a^2}{1}=\frac{b^2}{\sqrt{2}}=\frac{c}{1}\)

vậy min P =0

sorry bạn mình ko tìm đc giá trị lớn nhất mà chỉ tìm đc giá trị nhỏ nhất thôi

Do \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\a+b+c=5\end{matrix}\right.\) \(\Rightarrow c\le4\)

\(\Rightarrow2\le c\le4\Rightarrow\left(c-2\right)\left(c-4\right)\le0\Rightarrow c^2\le6c-8\)

\(0\le a\le1< 6\Rightarrow a\left(a-6\right)\le0\Rightarrow a^2\le6a\)

\(1\le b\le2< 5\Rightarrow\left(b-1\right)\left(b-5\right)\le0\Rightarrow b^2\le6b-5\)

Cộng vế:

\(a^2+b^2+c^2\le6\left(a+b+c\right)-13=17\)

\(A_{max}=17\) khi \(\left(a;b;c\right)=\left(0;1;4\right)\)

là -2 đấy

Đặt \(\left(a+1;b+1;c+1\right)=\left(x;y;z\right)\Rightarrow1\le x\le y\le z\le2\)

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

Do \(x\le y\le z\Rightarrow\left(z-y\right)\left(y-x\right)\ge0\)

\(\Leftrightarrow xy+yz\ge y^2+zx\)

\(\Leftrightarrow\dfrac{x}{z}+1\ge\dfrac{y}{z}+\dfrac{x}{y}\)

Tương tự: \(1+\dfrac{z}{x}\ge\dfrac{y}{x}+\dfrac{z}{y}\)

Cộng vế: \(2+\dfrac{x}{z}+\dfrac{z}{x}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{y}{x}\) (2)

Từ (1); (2) \(\Rightarrow B\le2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+5\)

Đặt \(\dfrac{z}{x}=t\Rightarrow1\le t\le2\)

\(\Rightarrow B\le2\left(t+\dfrac{1}{t}\right)+5=\dfrac{2t^2+2}{t}+5=\dfrac{2t^2+2}{t}-5+10\)

\(\Rightarrow B\le\dfrac{2t^2-5t+2}{t}+10=\dfrac{\left(t-2\right)\left(2t-1\right)}{t}+10\le10\)

\(B_{max}=10\) khi \(t=2\) hay \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\)

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