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\(\dfrac{x^2}{y+z}+\dfrac{1}{4}\left(y+z\right)\ge2.\sqrt{\dfrac{x^2}{y+z}.\dfrac{1}{4}\left(y+z\right)}=x\)
Tung tu : \(\dfrac{y^2}{x+z}+\dfrac{1}{4}\left(x+z\right)\ge y\)
\(\dfrac{z^2}{x+y}+\dfrac{1}{4}\left(x+y\right)\ge z\)
=> P+\(\dfrac{1}{4}\left(y+z\right)+\dfrac{1}{4}\left(x+z\right)+\dfrac{1}{4}\left(x+y\right)\ge x+y+z\)
=> P+\(\dfrac{1}{4}\left(2x+2y+2z\right)\ge4\)
=> P+2≥4
=> P≥2
Dau = khi: x=y=z=\(\dfrac{4}{3}\)
Vậy Min P=2 khi x=y=z=\(\dfrac{4}{3}\)
\(P=\sqrt[]{x}+\dfrac{3}{\sqrt[]{x}-1}\left(x>1\right)\)
\(P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\)
Áp dụng bất đẳng thức Cauchy cho 2 số \(\sqrt[]{x}-1;\dfrac{3}{\sqrt[]{x}-1}\) ta được :
\(\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{\sqrt[]{x}-1.\dfrac{3}{\sqrt[]{x}-1}}\)
\(\Rightarrow\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{3}\)
\(\Rightarrow P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\ge2\sqrt[]{3}+1\)
\(\Rightarrow Min\left(P\right)=2\sqrt[]{3}+1\)
\(\frac{4x + y}{xy} = \frac{4x}{xy} + \frac{y}{xy} = \frac{4}{y} + \frac{1}{x}\)
\(P = \frac{1}{x} + \frac{4}{y} + \frac{2x - y}{4}\)
\(=\frac{1}{x}+\frac{4}{y}+\frac{x}{2}-\frac{y}{4}\)
Theo BĐT Cauchy, ta có:
\(\frac{1}{x} + \frac{x}{4} \ge 2\sqrt{\frac{1}{x} \cdot \frac{x}{4}} = 2 \cdot \frac{1}{2} = 1\)
\(\frac{4}{y} + \frac{4y}{9} \ge 2\sqrt{\frac{4}{y} \cdot \frac{4y}{9}} = 2 \cdot \frac{4}{3} = \frac{8}{3}\)
\(P = \left( \frac{1}{x} + \frac{x}{4} \right) + \left( \frac{4}{y} + \frac{4y}{9} \right) + \left( \frac{x}{2} - \frac{x}{4} \right) - \frac{y}{4} - \frac{4y}{9}\)
\(=\left(\frac{1}{x}+\frac{x}{4}\right)+\left(\frac{4}{y}+\frac{4y}{9}\right)+\frac{x}{4}-\frac{25y}{36}\)
x+y=5
=>y=5-x
\(\frac{x}{4} - \frac{25y}{36} = \frac{5 - y}{4} - \frac{25y}{36} = \frac{5}{4} - \frac{y}{4} - \frac{25y}{36} = \frac{5}{4} - \frac{34y}{36} = \frac{5}{4} - \frac{17y}{18}\)
=>\(P \ge 1 + \frac{8}{3} + \frac{5}{4} - \frac{17y}{18}\)
Chọn điểm rơi tại y=3
Khi đó, ta sẽ có: \(P\ge1+\frac{8}{3}+\frac{5}{4}-\frac{17 \cdot3}{18}=\frac{25}{12}\)
Dấu '=' xảy ra khi y=3 và x=5-y=5-3=2
Ta có:
\(E\: =x^2+\frac{2x}{y}+\frac{1}{y^2}+y^2+\frac{2y}{x}+\frac{1}{x^2}=\left(x^2+y^2\right)+2\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\Rightarrow E\ge4+4+\frac{1}{x^2}+\frac{1}{y^2}=8+\frac{x^2+y^2}{x^2y^2}\)
Do: \(4=x^2+y^2\ge2xy\Rightarrow xy\le2\Rightarrow x^2y^2\le4\Rightarrow\frac{4}{x^2y^2}\ge1\)
\(\Rightarrow E\ge8+1=9\)
Dấu bằng xảy ra khi x=y=\(\sqrt{2}\)
\(2=x+y\ge2\sqrt{xy}\Rightarrow xy\le1.\)
\(\left(x^4+1\right)\left(y^4+1\right)+2013\ge2x^2.2y^2+2013\ge4+2013=2017\)
Min=2017
Dấu "=" xảy ra khi x=y=1
Lần sau tìm nơi gõ công thức và gõ hẳn ra nhé e <3
Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=x^4+y^4\ge\frac{\left(x^2+y^2\right)^2}{2}\ge\frac{\left(\frac{\left(x+y\right)^2}{2}\right)^2}{2}=\frac{\left(\frac{2^2}{2}\right)^2}{2}=...\text{(tự tính nhé :)}\)
Khi \(x=y=1\)
I spring. Because spring has many beautiful flowers.