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\(M=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\ge\dfrac{2}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{8}{\left(x+y\right)^2}=8\)
\(\Rightarrow M_{min}=8\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Câu 1:
Áp dụng BĐT Cô-si:
\(x^4+y^2\geq 2\sqrt{x^4y^2}=2x^2y\Rightarrow \frac{x}{x^4+y^2}\leq \frac{x}{2x^2y}=\frac{1}{2xy}=\frac{1}{2}(1)\)
\(x^2+y^4\geq 2\sqrt{x^2y^4}=2xy^2\Rightarrow \frac{y}{x^2+y^4}\leq \frac{y}{2xy^2}=\frac{1}{2xy}=\frac{1}{2}(2)\)
Lấy \((1)+(2)\Rightarrow A\leq \frac{1}{2}+\frac{1}{2}=1\)
Vậy \(A_{\max}=1\). Dấu bằng xảy ra khi \(x=y=1\)
Câu 2:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)(x^2+y^2+2xy)\geq (1+1)^2\)
\(\Rightarrow \frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{x^2+y^2+2xy}=\frac{4}{(x+y)^2}\geq \frac{4}{1}=4(*)\)
(do \(x+y\leq 1\) )
Áp dụng BĐT Cô-si:
\(\frac{1}{4xy}+4xy\geq 2\sqrt{\frac{4xy}{4xy}}=2(**)\)
\(x+y\geq 2\sqrt{xy}\Leftrightarrow 1\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}\)
\(\Rightarrow \frac{5}{4xy}\geq \frac{5}{4.\frac{1}{4}}=5(***)\)
Cộng \((*)+(**)+(***)\Rightarrow B\geq 4+2+5=11\)
Vậy \(B_{\min}=11\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
\(P=\dfrac{x}{y+1}+\dfrac{y}{x+1}=\dfrac{x}{2-x}+\dfrac{y}{2-y}\)
\(P+2=\dfrac{x}{2-x}+1+\dfrac{y}{2-y}+1=\dfrac{x+2-x}{2-x}+\dfrac{y+2-y}{2-y}\)
\(\dfrac{P+2}{2}=\dfrac{1}{2-x}+\dfrac{1}{2-y}\ge\dfrac{4}{2-x+2-y}=\dfrac{4}{4-\left(x+y\right)}=\dfrac{4}{4-1}=\dfrac{4}{3}\)
\(\dfrac{P+2}{2}\ge\dfrac{4}{3};P+2\ge\dfrac{8}{3};P\ge\dfrac{2}{3}\)
x=y=1/2
\(B=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\ge\frac{9}{2x+y+z+x+2y+z+x+y+2z}=\frac{9}{4\left(x+y+z\right)}\ge\frac{9}{4}.1=\frac{9}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
\(A\ge\frac{9}{2x+y+2y+z+2z+x}=\frac{9}{3\left(x+y+z\right)}=\frac{9}{3.3}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Cho x,y >=0 và x+y=1
Tìm GTNN của A= \(\left(\dfrac{x+1}{y}\right)^2+\left(\dfrac{y+1}{x}\right)^2\)
\(A=\left(\dfrac{x+1}{y}\right)^2+\left(\dfrac{y+1}{x}\right)^2\)
\(A=\left(\dfrac{x+x+y}{y}\right)^2+\left(\dfrac{y+x+y}{x}\right)^2\)
\(A=\left(\dfrac{2x}{y}+1\right)^2+\left(\dfrac{2y}{x}+1\right)^2\)
\(A=\dfrac{4x^2}{y^2}+\dfrac{4x}{y}+1+\dfrac{4y^2}{x^2}+\dfrac{4y}{x}+1\)
\(A\ge8+8+2=18\)
\(\Rightarrow MINA=18\Leftrightarrow x=y=\dfrac{1}{2}\)
PP : biến đổi tương đương
Bài làm
Ta có \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\)
\(\Leftrightarrow\dfrac{\left(x+y\right)\left(y+x\right)}{xy\left(x+y\right)}\ge\dfrac{4xy}{\left(x+y\right)xy}\)
Vì x , y >0 , ta suy ra (x+y)2 \(\ge\)4xy
\(\Leftrightarrow\left(x+y\right)^2-4xy\ge0\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
Hay (x-y)2 \(\ge\)0 ( điều này luôn đúng )
Vậy..........
chat lop 8.
x+y=1
(x-y)^2 ≥0
x^2+y^2-2xy ≥0
x^2+y^2≥2xy
x^2+y^2+2xy≥2xy+2xy
(x+y)^2≥4xy
1≥4xy
xy≤1/4
x,y>0=>xy>0
<=>1/xy≥4
(x+y)/xy≥4 ™#{1=x+y}!
1/y+1/x≥4
1/x+1/y≥4
Áp dụng BĐT Cô - si dạng Engel , ta có :
\(A=\dfrac{1}{x}+\dfrac{1}{y}\) ≥ \(\dfrac{\left(1+1\right)^2}{x+y}=\dfrac{4}{1}=4\)
⇒ AMIN = 4 ⇔ x = y = \(\dfrac{1}{2}\)