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\(P=\sqrt{\frac{1}{36}\left(11a+7b\right)^2+\frac{59\left(a-b\right)^2}{36}}+\sqrt{\frac{1}{36}\left(7a+11b\right)+\frac{59\left(a-b\right)^2}{36}}\)
\(=\sqrt{\frac{1}{16}\left(3a+5b\right)^2+\frac{5\left(a-b\right)^2}{16}}+\sqrt{\frac{1}{16}\left(5a+3b\right)^2+\frac{5\left(a-b\right)^2}{16}}\)
\(\ge\frac{1}{6}\left(11a+7b\right)+\frac{1}{6}\left(7a+11b\right)+\frac{1}{4}\left(3a+5b\right)+\frac{1}{4}\left(5a+3b\right)\)
\(=5\left(a+b\right)=5.2016=10080\)
Từ điều kiện suy ra \(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)
Áp dụng BĐT Cô-si, ta có :
\(3\le\sqrt{xy}+\sqrt{x}.1+\sqrt{y}.1\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}\)
\(\Rightarrow x+y\ge2\)
Ta có : \(\frac{x^2}{y}+y\ge2\sqrt{\frac{x^2}{y}.y}=2x\); \(\frac{y^2}{x}+x\ge2\sqrt{\frac{y^2}{x}.x}=2y\)
\(\Rightarrow\frac{x^2}{y}+\frac{y^2}{x}+x+y\ge2x+2y\)
\(\Rightarrow P=\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\ge2\)
Vậy GTNN của P là 2 khi x = y = 1
a/ Sửa đề:
\(\sqrt{22x^2+36xy+6y^2}+\sqrt{22y^2+36xy+6x^2}=x^2+y^2+32\)
\(\Leftrightarrow64x^2+64y^2+2048-64\sqrt{22x^2+36xy+6y^2}-64\sqrt{22y^2+36xy+6x^2}=0\)
\(\Leftrightarrow\left(22x^2+36xy+6y^2-64\sqrt{22x^2+36xy+6y^2}+1024\right)+\left(22y^2+36xy+6x^2-64\sqrt{22y^2+36xy+6x^2}+1024\right)+\left(36x^2-72xy+36y^2\right)=0\)
\(\Leftrightarrow\left(\sqrt{22x^2+36xy+y^2}-32\right)^2+\left(\sqrt{22y^2+36xy+6x^2}-32\right)^2+36\left(x-y\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{22x^2+36xy+6y^2}=32\\\sqrt{22y^2+36xy+6x^2}=32\\x=y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{64x^2}=32\\x=y\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=y=4\\x=y=-4\end{cases}}\)
Bài 1: Áp dụng BĐT AM-GM ta có:
\(1+x\ge2\sqrt{x}\)
\(x+y\ge2\sqrt{xy}\)
\(y+1\ge2\sqrt{y}\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{xy}+\sqrt{y}\right)\)
\(1+x+y\ge\sqrt{x}+\sqrt{xy}+\sqrt{y}\Leftrightarrow VT\ge VP\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\y+1=2\sqrt{y}\end{cases}}\Rightarrow x=y=1\)
Khi đó \(S=x^{2013}+y^{2013}=1^{2013}+1^{2013}=2\)
Bài 2: Vì \(\hept{\begin{cases}x,y,z\in\left[-1;3\right]\\x+y+z=3\end{cases}}\) nên
\(0\le\left(x+1\right)\left(y+1\right)\left(z+1\right)+\left(3-x\right)\left(3-y\right)\left(3-z\right)\)
\(\Leftrightarrow0\le4\left(xy+yz+xz\right)-8\left(x+y+z\right)+28\)
\(\Leftrightarrow0\le2\left(xy+yz+xz\right)+2\)
\(\Leftrightarrow x^2+y^2+z^2\le x^2+y^2+z^2+2\left(xy+yz+xz\right)+2\)
\(\Leftrightarrow x^2+y^2+z^2\le\left(x+y+z\right)^2+2\)
\(\Leftrightarrow x^2+y^2+z^2\le3^2+2=9+2=11\)
Dự đoán \(x=y=z=1\) ta tính được \(A=6+3\sqrt{2}\)
Ta sẽ c/m nó là GTLN của A
Thật vậy, ta cần chứng minh \(Σ\left(2+\sqrt{2}-2\sqrt{x}-\sqrt{1+x^2}\right)\ge0\)
\(\LeftrightarrowΣ\left(\frac{2\left(1-x\right)}{1+\sqrt{x}}+\frac{1-x^2}{\sqrt{2}+\sqrt{1+x^2}}\right)\ge0\)
\(\LeftrightarrowΣ\left(x-1\right)\left(1+\frac{1}{\sqrt{2}}-\frac{2}{1+\sqrt{x}}-\frac{x+1}{\sqrt{2}+\sqrt{1+x^2}}\right)+\left(1+\frac{1}{\sqrt{2}}\right)\left(3-x-y-z\right)\ge0\)
\(\LeftrightarrowΣ\left(x-1\right)^2\left(\frac{1}{\left(1+\sqrt{x}\right)^2}-\frac{x+1}{\sqrt{2}\left(\sqrt{2}+\sqrt{1+x^2}\right)\left(\sqrt{2}x+\sqrt{1+x^2}\right)}\right)+\left(1+\frac{1}{\sqrt{2}}\right)\left(3-x-y-z\right)\ge0\)
BĐT cuối đủ để chứng minh
\(\sqrt{2}\left(\sqrt{2}+\sqrt{1+x^2}\right)\left(\sqrt{2}x+\sqrt{1+x^2}\right)\ge\left(x+1\right)\left(1+\sqrt{x}\right)^2\)
Đặt \(1+x=2k\sqrt{x}\). Hence, theo Cauchy-Schwarz:
\(\sqrt{2}\left(\sqrt{2}+\sqrt{1+x^2}\right)\left(\sqrt{2}x+\sqrt{1+x^2}\right)\)
\(=\sqrt{2}\left(\sqrt{2}+\frac{1}{\sqrt{2}}\sqrt{2\left(1+x^2\right)}\right)\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\sqrt{2\left(1+x^2\right)}\right)\)
\(\ge\sqrt{2}\left(\sqrt{2}+\frac{x+1}{\sqrt{2}}\right)\left(\sqrt{2}x+\frac{x+1}{\sqrt{2}}\right)\)
\(=\frac{1}{\sqrt{2}}\left(x+3\right)\left(3x+1\right)=\frac{1}{\sqrt{2}}\left(3x^2+10x+3\right)\)
\(=\frac{1}{\sqrt{2}}\left(3\left(4k^2-2\right)x+10x\right)2\sqrt{2}x\left(3k^2+1\right)\)
Mặt khác \(\left(x+1\right)\left(1+\sqrt{x}\right)^2=\left(x+1\right)\left(x+1+2\sqrt{x}\right)\)
\(=2k\left(2k+2\right)x=4k\left(k+1\right)x\). Có nghĩa là ta cần phải c/m
\(3k^2+1\ge\sqrt{2}k\left(k+1\right)\Leftrightarrow\left(3-\sqrt{2}\right)k^2-2\sqrt{k}+1\ge0\)
Nó đúng theo AM-GM
\(\left(3-\sqrt{2}\right)k^2-\sqrt{2}k+1\ge\left(2\sqrt{3-\sqrt{2}}-\sqrt{2}\right)k\ge0\)
Hơi đẹp nhỉ nhưng xong r` đó :D
bunyakovsky:
\(\left(\sqrt{1+x^2}+\sqrt{2x}\right)^2\le2\left(x+1\right)^2\)
\(\Leftrightarrow\sqrt{1+x^2}+\sqrt{2}.\sqrt{x}\le\sqrt{2}\left(x+1\right)\)
tương tự :phần còn lại + thêm với\(\left(2-\sqrt{2}\right)\left(x+y+z\right)\)