Câu 1: Rút gọn biểu thức
a) \(N=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
b) \(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
Câu 2:
a) Cho a > 0. Chứng minh: \(a+\dfrac{1}{a}\ge2\)
b) Cho \(a\ge0\) , \(b\ge0\) . Chứng minh: \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
c) Cho a, b > 0. Chứng minh: \(\sqrt{a}+\sqrt{b}\le\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\)
d) Chứng minh: \(\dfrac{a^2+2}{\sqrt{a^2+1}}\ge2\) với mọi a
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:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
2b)
Biến đổi tương đương:
\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\) (1)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\dfrac{a+2\sqrt{ab}+b}{4}\)
\(\Leftrightarrow2a+2b\ge a+2\sqrt{ab}+b\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) luôn đúng
=> (1) đúng
Dấu "=" xảy ra khi a = b.
2c)
Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:
\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\) (đpcm)
Dấu "=" xảy ra khi a = b.
2d)
Áp dụng BĐT AM - GM, ta có:
\(\dfrac{a^2+2}{\sqrt{a^2+1}}=\dfrac{a^2+1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\dfrac{1}{\sqrt{a^2+1}}\ge2\) (đpcm)
Dấu "=" xảy ra khi a = 0
Bài 2:
\(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Leftrightarrow\dfrac{\left(a^2-2a+1\right)+2a}{a}\ge2\)
\(\Leftrightarrow\dfrac{\left(a-1\right)^2+2a}{a}\ge2\)
\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a}+2\ge2\)
\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a}\ge0\)
Vì \(a>0\Rightarrow\dfrac{\left(a+1\right)^2}{a}\ge0\Rightarrowđpcm\)
b, \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)
\(\Leftrightarrow\dfrac{a+b}{2}-\dfrac{a+2\sqrt{ab}+b}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a+2b-a-2\sqrt{ab}-b}{4}\ge0\)
\(\Leftrightarrow\dfrac{a-2\sqrt{ab}+b}{a}\ge0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{4}\ge0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{a}-\sqrt{b}}{2}\right)^2\ge0\)
Đúng với \(a\ge0;b\ge0\)
d, \(\dfrac{a^2+2}{\sqrt{a^2+1}}\)
Ta có:
\(a^2+2=\left(\sqrt{a^2+1}\right)^2+1\) nên
\(\Rightarrow\dfrac{a^2+2}{\sqrt{a^2+1}}=\dfrac{\left(\sqrt{a^2+1}\right)^2+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\dfrac{1}{\sqrt{a^2+1}}\ge2\)( chứng minh phần a)
Áp Cauchy vô
Vậy cx hỏi