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Đặt VT là T
Áp dụng AM-GM cho 3 số dương, ta có:
\(\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)
\(T\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)(đpcm)
\(P=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{.....}+\dfrac{x+2}{....}\)
\(=\dfrac{\sqrt{x^3}+2x+2\sqrt{x}-2+x+2}{.....}=\dfrac{\sqrt{x^3}+3x+2\sqrt{x}}{....}\)
\(=\dfrac{\sqrt{x}\left(x+3\sqrt{x}+2\right)}{....}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{....}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P/S: Chú ý điều kiện khi rút gọn, tự tìm.
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
Bài 1:
\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )
Khi đó:
\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)
\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)
(Áp dụng BĐT Cauchy_Schwarz)
Theo BĐT Cauchy dễ thấy:
\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$
Bài 2:
Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)
Ta có:
\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)
\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)
Áp dụng BĐT Cauchy:
\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)
\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)
\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)
Nhân theo vế:
\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)
\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)
\(\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$
Câu hỏi của Vo Trong Duy - Toán lớp 9 - Học toán với OnlineMath
Nguyễn Huy Thắng cái dòng thứ 3 từ dưới lên hơi khó hiểu
\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=8\)
=>\(8xyz=xyz+\sum x+\sum xy+1\)
=>\(\sum x^2+14xyz=\left(\sum x\right)^2+2\sum x+2\)
mặt khác
\(8=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge\dfrac{8}{\sqrt[3]{xyz}}\rightarrow xyz\ge1\)
đặt \(\sum x=a\left(a\ge3\right)\)
khi đó \(P=\dfrac{a^2+2a+2}{4a^2+15xyz}\le\dfrac{a^2+2a+2}{4a^2+15}\)
\(\dfrac{a^2+2a+2}{4a^2+15}=\dfrac{1}{3}-\dfrac{\left(a-3\right)^2}{12a^2+45}\le\dfrac{1}{3}\)
vậy max bằng 1/3 khi x=y=z=1
A=\(1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{y}{x}\)
A= \(\left(x+\dfrac{1}{2x}\right)+\left(y+\dfrac{1}{2y}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+2\)
Áp Dụng BĐT Cô si ta có:
\(\left(x+\dfrac{1}{2x}\right)\ge\sqrt{2}\); \(\left(y+\dfrac{1}{2y}\right)\ge\sqrt{2}\); \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)
\(\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge2\sqrt{\dfrac{1}{2x.2y}}=\dfrac{1}{\sqrt{xy}}\ge\dfrac{\sqrt{2}}{\sqrt{x^2+y^2}}=\sqrt{2}\)
suy ra A\(\ge4+3\sqrt{2}\)
Dấu = xảy ra
\(\left\{{}\begin{matrix}x=y\\x=\dfrac{1}{2x}\\y=\dfrac{1}{2y}\end{matrix}\right.\)
\(\Leftrightarrow\)x=y=\(\dfrac{\sqrt{2}}{2}\)
Vậy Min A=4+3\(\sqrt{2}\) khi x=y=\(\dfrac{\sqrt{2}}{2}\)
Trước hết ta có \(\dfrac{\left(x+y\right)^2}{2}\le x^2+y^2\Rightarrow x+y\le\sqrt{2\left(x^2+y^2\right)}=\sqrt{2}\)
\(A=1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{y}{x}\)
\(A=2+x+y+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{x}{y}+\dfrac{y}{x}\ge2+x+y+\dfrac{4}{x+y}+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}\)
\(\Rightarrow A\ge4+x+y+\dfrac{4}{x+y}=4+x+y+\dfrac{2}{x+y}+\dfrac{2}{x+y}\)
\(\Rightarrow A\ge4+2\sqrt{\left(x+y\right).\dfrac{2}{\left(x+y\right)}}+\dfrac{2}{\sqrt{2}}=4+3\sqrt{2}\)
\(\Rightarrow A_{min}=4+3\sqrt{2}\) khi \(x=y=\dfrac{1}{\sqrt{2}}\)
cho biểu thức trên = P
\(P=\left(xy\right)^2+\dfrac{1}{\left(xy\right)^2}+2=256\left(xy\right)^2+\dfrac{1}{\left(xy\right)^2}+2-255\left(xy\right)^2< =>P\ge34-255\left(xy\right)^2\)
ta lại có \(x+y\ge2\sqrt{xy}=>1\ge2\sqrt{xy}=>\dfrac{1}{16}\ge\left(xy\right)^2\)
=> \(P\ge34-\dfrac{255}{16}=18\dfrac{1}{16}\)
Dấu = xảy ra khi x=y=1/2