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Theo đề thì:\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)
\(\Leftrightarrow xz+yz-2xy=0\)
Cũng từ \(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)
\(\Leftrightarrow\dfrac{2}{z}=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)
\(\Leftrightarrow z\le\sqrt{xy}\)
\(\Leftrightarrow z^2\le xy\)
Quay lại bài toán ta có:
\(T=\dfrac{x+z}{2x-z}+\dfrac{z+y}{2y-z}=\dfrac{2z^2-6xy-\left(xz+yz-2xy\right)}{-z^2+2\left(xz+yz-2xy\right)}\)
\(=\dfrac{6xy-2z^2}{z^2}\ge\dfrac{6xy-2xy}{xy}=4\)
Vậy GTNN là T = 4 khi x = y = z = 1
ta có : \(\left(x+y-1\right)^2=xy\Leftrightarrow x^2+y^2+xy-2x-2y+1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+xy-1=0\)
\(0=\left(x-1\right)^2+\left(y-1\right)^2+xy-1\ge xy-1\)
\(\Leftrightarrow xy\le1\)
mà \(xy=\left(x+y-1\right)^2\le1\Leftrightarrow-1\le x+y-1\le1\)
\(\Leftrightarrow0\le x+y\le2\).
\(VT=\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{\sqrt{xy}}{x+y}\)
Áp dụng bất đẳng thức cauchy dạng phân thức:
\(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\ge\dfrac{4}{\left(x+y\right)^2}\ge\dfrac{4}{4}=1\)(*)
vì \(xy\le1\)nên \(\sqrt{xy}\ge xy\)( đúng vì nó tương đương \(\sqrt{xy}\left(1-\sqrt{xy}\right)\ge0\))
\(\Rightarrow\dfrac{1}{2xy}+\dfrac{\sqrt{xy}}{x+y}\ge\dfrac{1}{2\sqrt{xy}}+\dfrac{\sqrt{xy}}{2}\)( vì \(x+y\le2\))
Áp dụng bất đẳng thức cauchy: \(\dfrac{1}{2\sqrt{xy}}+\dfrac{\sqrt{xy}}{2}\ge2\sqrt{\dfrac{1}{2\sqrt{xy}}.\dfrac{\sqrt{xy}}{2}}=1\)(**)
từ (*) và (**) ta có \(VT\ge1+1=2\)
đẳng thức xảy ra khi x=y=1
mình k ghi lại đề nữa ta có
\(1\ge\dfrac{4^2}{x+24}+\dfrac{5^2}{y+16}+\dfrac{3^2}{z+4}\ge\dfrac{\left(4+5+3\right)^2}{x+y+z+24+16+4}=\dfrac{12^2}{x+y+z+44}\)
=>x+y+z+44>=12^2=144=> x+y+z=100
đặt x+y+z=a(a>=100)
\(x+y+z+\dfrac{1}{x+y+z}=a+\dfrac{1}{a}=\dfrac{a}{10000}+\dfrac{1}{a}+\dfrac{9999a}{10000}\ge\dfrac{2}{100}+\dfrac{9999a}{10000}\)
do a>=100 nên
\(a+\dfrac{1}{a}\ge\dfrac{2}{100}+\dfrac{9999}{100}=\dfrac{10001}{100}\) khi a= 100 hay x+y+z=100
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
\(P=\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}=\dfrac{1}{2021}.\dfrac{2022^2}{\dfrac{2022}{2021}}=2022\)
\(P_{min}=2022\) khi \(\left(x;y\right)=\left(1;\dfrac{1}{2021}\right)\)
sao cái đoạn \(\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}\) làm kiểu gì ra thầy :)