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1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
a, \(y=\dfrac{\sqrt{x-2}}{x}=\sqrt{\dfrac{1}{x}-\dfrac{2}{x^2}}\ge0\)
\(min=0\Leftrightarrow\dfrac{1}{x}-\dfrac{2}{x^2}=0\Leftrightarrow x=2\)
b, Áp dụng BĐT Cosi:
\(f\left(x\right)=\dfrac{x}{\sqrt{x-1}}=\dfrac{x-1+1}{\sqrt{x-1}}=\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\ge2\)
\(minf\left(x\right)=2\Leftrightarrow x=2\)
Bạn tham khảo:
cho x,y,z >0 thỏa mãn \(2\sqrt{y}+\sqrt{z}=\dfrac{1}{\sqrt{x}}\). CMR: \(\dfrac{3yz}{x}+\dfrac{4zx}{y}+\dfrac{5xy}{z}\ge... - Hoc24
1. 1/x + 2/1-x = (1/x - 1) + (2/1-x - 2) + 3
= 1-x/x + (2-2(1-x))/1-x + 3
= 1-x/x + 2x/1-x + 3 >= 2√2 + 3
Dấu "=" xảy ra khi x =√2 - 1
2. a = √z-1, b = √x-2, c = √y-3 (a,b,c >=0)
=> P = √z-1 / z + √x-2 / x + √y-3 / y
= a/a^2+1 + b/b^2+2 + c/c^2+3
a^2+1 >= 2a => a/a^2+1 <= 1/2
b^2+2 >= 2√2 b => b/b^2+2 <= 1/2√2
c^2+3 >= 2√3 c => c/c^2+3 <= 1/2√3
=> P <= 1/2 + 1/2√2 + 1/2√3
Dấu = xảy ra khi a^2 = 1, b^2 = 2, c^2 =3
<=> z-1 = 1, x-2 = 2, y-3 = 3
<=> x=4, y=6, z=2
\(P=\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{xy+xz+yz}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{1}{x}+\dfrac{y^2}{2}+\dfrac{1}{y}+\dfrac{z^2}{2}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\left(\dfrac{x^2}{2}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)+\left(\dfrac{y^2}{2}+\dfrac{1}{2y}+\dfrac{1}{2y}\right)+\left(\dfrac{z^2}{2}+\dfrac{1}{2z}+\dfrac{1}{2z}\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{x^2}{2}.\dfrac{1}{2x}.\dfrac{1}{2x}}+3\sqrt[3]{\dfrac{y^2}{2}.\dfrac{1}{2y}.\dfrac{1}{2y}}+3\sqrt[3]{\dfrac{z^2}{2}.\dfrac{1}{2z}.\dfrac{1}{2z}}=\dfrac{9}{2}\)
\(\Rightarrow P_{min}=\dfrac{9}{2}\) khi \(x=y=z=1\)
\(P=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\Rightarrow P^2=\dfrac{x^2}{y}+\dfrac{y^2}{x}+2\sqrt{xy}\)
\(P^2=\left(\dfrac{x^2}{y}+\sqrt{xy}+\sqrt{xy}\right)+\left(\dfrac{y^2}{x}+\sqrt{xy}+\sqrt{xy}\right)-2\sqrt{xy}\)
\(P^2\ge3x+3y-2\sqrt{xy}\ge3\left(x+y\right)-\left(x+y\right)=2\left(x+y\right)=4038\)
\(\Rightarrow P\ge\sqrt{4038}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{2019}{2}\)
Ta có:
\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{y-2019}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\ge\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
Lại có:
\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{2019-y}}=\dfrac{2019-y}{\sqrt{y}}+\dfrac{2019-x}{\sqrt{x}}\\ =\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}-\sqrt{x}-\sqrt{y}\)
\(\Rightarrow2P=\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}=2019\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\ge2019\cdot\dfrac{2}{\sqrt[4]{xy}}\\ \ge2019\dfrac{2}{\sqrt[2]{\dfrac{x+y}{2}}}=2019\cdot\dfrac{2}{\sqrt{\dfrac{2019}{2}}}=2\sqrt{2}\sqrt{2019}\)
\(\Rightarrow P\ge\sqrt{2}\sqrt{2019}\)
Dấu = khi \(x=y=\dfrac{2019}{2}\)
\(y=\dfrac{x-1}{2}+\dfrac{1}{2}+\dfrac{2}{x-1}\ge2\sqrt{\dfrac{x-1}{2}\cdot\dfrac{2}{x-1}}+\dfrac{1}{2}=2\cdot1+\dfrac{1}{2}=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow\left(x-1\right)^2=2\Leftrightarrow x=3\left(x>1\right)\)
Lời giải:
$x>1\Rightarrow x-1>0$
Áp dụng BĐT Cô-si ta có:
$y=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\geq 2\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=2+\frac{1}{2}=\frac{5}{2}$
Vậy $y_{\min}=\frac{5}{2}$
Giá trị này đạt tại $x-1=2\Leftrightarrow x=3$