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2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
điều kiện xác định : \(a>0\)
ta có : \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{a^2-\sqrt{a}}{a+\sqrt{a}+1}+\dfrac{1}{\sqrt{a}}\)
\(\Leftrightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(\sqrt{a}^3-1\right)}{a+\sqrt{a}+1}+\dfrac{1}{\sqrt{a}}\)
\(\Leftrightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}+\dfrac{1}{\sqrt{a}}\)\(\Leftrightarrow A=\sqrt{a}\left(\sqrt{a}+1\right)-\sqrt{a}\left(\sqrt{a}-1\right)+\dfrac{1}{\sqrt{a}}\)
\(\Leftrightarrow A=a+\sqrt{a}-a+\sqrt{a}+\dfrac{1}{\sqrt{a}}=2\sqrt{a}+\dfrac{1}{\sqrt{a}}\)
áp dụng bất đẳng thức cô si ta có : \(A=2\sqrt{a}+\dfrac{1}{\sqrt{a}}\ge2\sqrt{2}\Rightarrow\left(đpcm\right)\)
a) \(a+b-2\sqrt{ab}\ge0\)
<=> \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge0\) (luôn đúng )
=> đpcm
b) \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\Leftrightarrow\sqrt{\dfrac{a+b}{2}^2}\ge\left(\dfrac{\sqrt{a}+\sqrt{b}}{2}\right)^2\)
<=> \(\dfrac{a+b}{2}\ge\dfrac{a+b+2\sqrt{ab}}{4}\)
<=> \(\dfrac{2a+2b}{4}\ge\dfrac{a+b+2\sqrt{ab}}{4}\Leftrightarrow2a+2b\ge a+b+2\sqrt{ab}\)
<=> \(2a+2b-a-b-2\sqrt{ab}\ge0\)
<=> \(a-2\sqrt{ab}+b\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
=> đpcm
b)Áp dụng BĐT AM-GM ta có:
\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\sqrt{\dfrac{\sqrt{a}}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}}=2\)
Xảy ra khi \(a=b\)
c)Áp dụng BĐT \(x^2+y^2\ge2xy\) có:
\(VT=\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)
\(\ge2\sqrt{\left(a+b\right)\cdot2\sqrt{ab}}=2\sqrt{2\left(a+b\right)\cdot\sqrt{ab}}=VP\)
Xảy ra khi \(a=b\)
a)\(\dfrac{a^2+3}{\sqrt{a^2+3}}=\sqrt{a^2+3}\ge\sqrt{3}< 2\)\
sai đề
đk: a > 0
A/dụng bđt AM-GM có:
\(A=2\sqrt{a}+\dfrac{1}{\sqrt{a}}\ge2\sqrt{2\sqrt{a}\cdot\dfrac{1}{\sqrt{a}}}=2\sqrt{2\cdot1}=2\sqrt{2}\left(đpcm\right)\)
Đẳng thức xảy ra khi :\(2\sqrt{a}=\dfrac{1}{\sqrt{a}}\Leftrightarrow a=\dfrac{1}{2}\)