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2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)
\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{2xy}+\frac{\left(x+y\right)^2}{2xy}\)
\(S\ge\frac{4\left(x+y\right)^2}{x^2+y^2+2xy}+\frac{\left(x+y\right)^2}{\frac{\left(x+y\right)^2}{2}}=\frac{4\left(x+y\right)^2}{\left(x+y\right)^2}+2=6\)
\(\Rightarrow S_{min}=6\) khi \(x=y\)
\(\dfrac{1}{2xy}\ge\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=2\)
Dấu = xảy ra khi \(x=y=\dfrac{1}{2}\)
a) \(4xy\le\left(x+y\right)^2=1\)
=> \(xy\le4\)
Dấu "=" xảy ra <=> x = y = 1/2
b) A = \(A=x^2+2+\dfrac{1}{x^2}+y^2+2+\dfrac{1}{y^2}=\left(x^2+y^2\right)+\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\ge2xy+\dfrac{2}{xy}+4=\left(32xy+\dfrac{2}{xy}\right)-30xy+4\ge8-\dfrac{30}{4}+4=\dfrac{9}{2}\)
Dấu "=" xảy ra <=> x = y = 1/2
\(P=\dfrac{18}{x^2+y^2}+\dfrac{5}{xy}=\dfrac{18\left(x+y\right)^2}{x^2+y^2}+\dfrac{5\left(x+y\right)^2}{xy}=\dfrac{18\left[\left(x^2+y^2\right)+2xy\right]}{x^2+y^2}+\dfrac{5\left[\left(x^2+y^2\right)+2xy\right]}{xy}=18+\dfrac{36xy}{x^2+y^2}+\dfrac{5\left(x^2+y^2\right)}{xy}+10=28+\left[\dfrac{36xy}{x^2+y^2}+\dfrac{5\left(x^2+y^2\right)}{xy}\right]\overset{Cauchy}{\ge}28+2\sqrt{\dfrac{36xy}{x^2+y^2}.\dfrac{5\left(x^2+y^2\right)}{xy}}=28+2.6\sqrt{5}=28+12\sqrt{5}\)
=> \(P^{ }_{min}=28+12\sqrt{5}\) khi và chỉ khi \(\left\{{}\begin{matrix}\dfrac{36xy}{x^2+y^2}=\dfrac{5\left(x^2+y^2\right)}{xy}\\x+y=1\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5-\sqrt{5}}{4}\\y=\dfrac{\sqrt{5}-1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\y=\dfrac{5-\sqrt{5}}{4}\end{matrix}\right.\end{matrix}\right.\)
Lời giải:
Ta có \(B=\frac{x}{y}+\frac{y}{x}+\frac{xy}{x^2+xy+y^2}=\frac{8}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{xy}{x^2+xy+y^2}\)
\(=\frac{8}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{x^2+xy+y^2}{9xy}+\frac{xy}{x^2+xy+y^2}-\frac{1}{9}\)
Áp dụng BĐT AM-GM:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(\frac{x^2+xy+y^2}{9xy}+\frac{xy}{x^2+xy+y^2}\geq 2\sqrt{\frac{1}{9}}=\frac{2}{3}\)
Do đó: \(B\geq \frac{8}{9}.2+\frac{2}{3}-\frac{1}{9}=\frac{7}{3}\Leftrightarrow B_{\min}=\frac{7}{3}\)
Dấu bằng xảy ra khi $x=y$
\(\dfrac{4}{2xy}+\dfrac{4}{x^2+y^2}-\dfrac{1}{x^2+y^2}\)
\(\ge\dfrac{\left(2+2\right)^2}{x^2+y^2+2xy}-\dfrac{1}{x^2+y^2}=16-\dfrac{1}{x^2+y^2}\)
\(=16-\dfrac{2}{2\left(x^2+y^2\right)}\ge\dfrac{16}{\left(x+y\right)^2}=14\)
Dấu = xảy ra khi \(x=y=\dfrac{1}{2}\)
Cách khác
Đặt xy=t
\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy=1-2t\)
\(\Rightarrow M=\dfrac{2}{t}+\dfrac{3}{1-2t}\)
\(M=\dfrac{2-4t+3t}{t-2t^2}=\dfrac{2-t}{t-2t^2}\)(đến đây dùng phương pháp delta)
giải chi tiết jumf mk vì mk chưa hok delta
pp delta để mình xem lại
Tới đó ta sẽ cm:\(M=\dfrac{2-t}{t-2t^2}\ge14\)
\(\Leftrightarrow2-t\ge14t-28t^2\)
\(\Leftrightarrow28t^2-15t+2\ge0\)
\(\Leftrightarrow28t^2-8t-7x+2\ge0\)
\(\Leftrightarrow28t\left(t-\dfrac{2}{7}\right)-7\left(t-\dfrac{2}{7}\right)\ge0\)
\(\Leftrightarrow7\left(4t-1\right)\left(t-\dfrac{2}{7}\right)\ge0\)(vì \(t=xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\))
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