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Mấy cái dấu "=" anh tự xét.
Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)
b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)
Áp dụng bất đẳng thức Cauchy :
\(\frac{x^4}{y^2\left(x+z\right)}+\frac{y^2}{2x}+\frac{x+z}{4}\ge3\sqrt[3]{\frac{x^4\cdot y^2\cdot\left(x+z\right)}{y^2\cdot\left(x+z\right)\cdot2x\cdot4}}=3\sqrt[3]{\frac{x^3}{8}}=\frac{3x}{2}\)
Tương tự ta cũng có :
\(\frac{y^4}{z^2\left(x+y\right)}+\frac{z^2}{2y}+\frac{x+y}{4}\ge\frac{3y}{2}\)
\(\frac{z^4}{x^2\left(y+z\right)}+\frac{x^2}{2z}+\frac{y+z}{4}\ge\frac{3z}{2}\)
Cộng theo vế ta được :
\(VT+\left(\frac{y^2}{2x}+\frac{z^2}{2y}+\frac{x^2}{2z}\right)+\frac{2\left(x+y+z\right)}{4}\ge\frac{3x}{2}+\frac{3y}{2}+\frac{3z}{2}\)
\(\Leftrightarrow VT+\frac{1}{2}\left(\frac{y^2}{x}+\frac{z^2}{y}+\frac{x^2}{z}\right)+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT+\frac{1}{2}\cdot\frac{\left(x+y+z\right)^2}{x+y+z}+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT+\frac{1}{2}\left(x+y+z\right)+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{x+y+z}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Từ bđt Cauchy : \(a+b\ge2\sqrt{ab}\) ta suy ra được \(ab\le\frac{\left(a+b\right)^2}{4}\)
Áp dụng vào bài toán của bạn :
a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{\left(x+3+5-x\right)^2}{4}=...............\)
b/ Tương tự
c/ \(y=\left(x+3\right)\left(5-2x\right)=\frac{1}{2}.\left(2x+6\right)\left(5-2x\right)\le\frac{1}{2}.\frac{\left(2x+6+5-2x\right)^2}{4}=.............\)
d/ Tương tự
e/ \(y=\left(6x+3\right)\left(5-2x\right)=3\left(2x+1\right)\left(5-2x\right)\le3.\frac{\left(2x+1+5-2x\right)^2}{4}=.......\)
f/ Xét \(\frac{1}{y}=\frac{x^2+2}{x}=x+\frac{2}{x}\ge2\sqrt{x.\frac{2}{x}}=2\sqrt{2}\)
Suy ra \(y\le\frac{1}{2\sqrt{2}}\)
..........................
g/ Đặt \(t=x^2\) , \(t>0\) (Vì nếu t = 0 thì y = 0)
\(\frac{1}{y}=\frac{t^3+6t^2+12t+8}{t}=t^2+6t+\frac{8}{t}+12\)
\(=t^2+6t+\frac{8}{3t}+\frac{8}{3t}+\frac{8}{3t}+12\)
\(\ge5.\sqrt[5]{t^2.6t.\left(\frac{8}{3t}\right)^3}+12=.................\)
Từ đó đảo ngược y lại rồi đổi dấu \(\ge\) thành \(\le\)
4.Áp dụng bđt Cô-si, tìm GTLN:
a)\(y=\frac{5x}{x^2+4};x>0\)
b)\(y=\frac{x^2}{\left(x^2+3\right)^3}\)
\(y=\frac{5x}{x^2+4}\le\frac{5x}{2\sqrt{x^2.4}}=\frac{5}{4}\)
Dấu "=" xảy ra khi \(x=2\)
\(y=\frac{x^2}{\left(x^2+\frac{3}{2}+\frac{3}{2}\right)^3}\le\frac{x^2}{\left(3\sqrt[3]{x^2.\frac{3}{2}.\frac{3}{2}}\right)^3}=\frac{4x^2}{243x^2}=\frac{4}{243}\)
Dấu "=" xảy ra khi \(x=\frac{\sqrt{6}}{2}\)
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y\right)^2+\frac{3}{\left(x+y\right)^2}+\left(x-y\right)^2=7\\x+y+\frac{1}{x+y}+x-y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y+\frac{1}{x+y}\right)^2+\left(x-y\right)^2=13\\x+y+\frac{1}{x+y}+x-y=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\frac{1}{x+y}=a\\x-y=b\end{matrix}\right.\) với \(\left|a\right|\ge2\)
\(\Rightarrow\left\{{}\begin{matrix}3a^2+b^2=13\\a+b=3\end{matrix}\right.\) \(\Rightarrow3a^2+\left(3-a\right)^2=13\)
\(\Leftrightarrow2a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=2\Rightarrow b=1\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+\frac{1}{x+y}=2\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2\left(x+y\right)+1=0\\x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-1\right)^2=0\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x-y=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
