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Theo cô-si thì \(2\sqrt{2x.3y}\le2x+3y\le2\Rightarrow xy\le\frac{1}{6}\)
\(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{26}{3xy}\)
\(\ge\frac{\left(2+2\right)^2}{4x^2+9y^2+12xy}+\frac{26}{\frac{3.1}{6}}\)
\(=\frac{14}{\left(2x+3y\right)^2}+\frac{26.6}{3}=56\)
\("="\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}\)
ta thấy \(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{26}{3xy}\ge\frac{16}{\left(2x+3y\right)^2}+\frac{26}{3xy}\)(1)
lại có \(2x+3y\le2\Leftrightarrow\left(2x+3y\right)^2\le4\Leftrightarrow4x^2+9y^2+12xy\le4\left(2\right)\)
mặt khác \(4x^2+9y^2\ge12xy\)(theo Bất Đẳng Thức Cosi cho x,y>0) (3)
từ (1) và (2) => \(12xy+12xy\le4\Leftrightarrow3xy\le\frac{1}{2}\left(4\right)\)
từ (1) và (4) => \(A\ge\frac{16}{4}+\frac{26}{\frac{1}{2}}=4+52=56\)
dấu "=" xảy ra khi \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}\)
Ta có:
\(\left(4x+9y+16z\right)\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(\sqrt{\frac{4x}{x}}+\sqrt{\frac{9y.25}{y}}+\sqrt{\frac{16z.64}{z}}\right)^2\)
\(\Leftrightarrow49\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(2+15+32\right)^2\)
\(\Leftrightarrow\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\ge49\)
Dấu = xảy ra tại \(x=\frac{1}{2};y=\frac{5}{3};z=2\)
a/ Cho x, y ≥ 1. Chứng minh: 1/(1 + x^2) + 1/(1 + y^2) ≥ 2/(1 + xy)
b/ Đề:...Tìm GTLN
Có:
\(\dfrac{1}{4x^2-4x+2}=\dfrac{1}{\left(2x-1\right)^2+1}\le\dfrac{1}{2}\forall x\ge1\)
\(\dfrac{1}{9y^2+6y+2}=\dfrac{1}{\left(3y+1\right)^2+1}\le\dfrac{1}{2}\forall y\ge0\)
\(\Rightarrow A=\dfrac{1}{4x^2-4x+2}+\dfrac{1}{9y^2+6y+2}\le\dfrac{1}{2}+\dfrac{1}{2}=1\)
Vậy MAXA = 1 khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
Áp dụng BĐT Cô-si ta có:
\(2x^2+3xy+4y^2\ge3\sqrt[3]{2x^2\cdot3xy\cdot4y^2}=3\sqrt[3]{24x^3y^3}\Rightarrow\sqrt{2x^2+3xy+4y^2}\ge\sqrt{xy\cdot3\sqrt[3]{24}}\)
Tương tự: \(\sqrt{2y^2+3yz+4z^2}\ge\sqrt{yz\cdot3\sqrt[3]{24}}\); \(\sqrt{2z^2+3zx+4x^2}\ge\sqrt{zx\cdot3\sqrt[3]{24}}\)
Cộng theo vế 3 BĐT vừa tìm, ta được:
\(P\ge\sqrt{3\sqrt[3]{24}}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\sqrt{3\sqrt[3]{24}}=\sqrt[6]{648}\)
Áp dụng bđt Svác xơ, ta có:
\(A\ge\dfrac{\left(\sqrt{2x}+\sqrt{3y}+\sqrt{4z}\right)^2}{2\left(4x^2+9y^2+16z^2\right)}\)\(=\dfrac{2x+3y+4z+2\left(\sqrt{6xy}+\sqrt{12yz}+\sqrt{8xz}\right)}{2}\)\(\ge\dfrac{1+2\left(3\sqrt[3]{\sqrt{576x^2y^2z^2}}\right)}{2}\)(BĐT Cô-si)\(\ge\dfrac{1+6}{2}=\dfrac{7}{2}\)
Vậy Amin=\(\dfrac{7}{2}\Leftrightarrow\)\(\left\{{}\begin{matrix}\dfrac{2x}{9y^2+16z^2}=\dfrac{3y}{4x^2+16z^2}=\dfrac{4z}{4x^2+9y^2}\\\sqrt{6xy}=\sqrt{12yz}=\sqrt{8xz}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}y=2z\)
Viết lại bài toán: Cho \(a^2+b^2+c^2=1\). Tìm max \(\sum\dfrac{a}{b^2+c^2}\)
với a=2x, b=3y, c=4z.
Áp dụng BĐT AM-GM:
\(a\left(b^2+c^2\right)=\dfrac{1}{\sqrt{2}}\sqrt{2a^2\left(1-a^2\right)\left(1-a^2\right)}\le\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{8}{27}}=\dfrac{2}{3\sqrt{3}}\)
Do đó \(VT\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
Vậy \(A_{Min}=\dfrac{3\sqrt{3}}{2}\)