Cho a>0,b>0,c>0. Chứng minh \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}\sqrt{\dfrac{c}{a+b}}\ge2\)
cho a,b,c > 0 thỏa mãn a + b + c = 6. Chứng minh:
\(\dfrac{a}{\sqrt{b^3+1}}+\dfrac{b}{\sqrt{c^3+1}}+\dfrac{c}{\sqrt{a^3+1}}\ge2\)
\(\dfrac{a}{\sqrt{b^3+1}}=\dfrac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\ge\dfrac{2a}{b+1+b^2-b+1}=\dfrac{2a}{b^2+2}\)
Tương tự và cộng lại:
\(VT\ge\dfrac{2a}{b^2+2}+\dfrac{2b}{c^2+2}+\dfrac{2c}{a^2+2}=a-\dfrac{ab^2}{b^2+2}+b-\dfrac{bc^2}{c^2+2}+c-\dfrac{ca^2}{a^2+2}\)
\(VT\ge6-\left(\dfrac{ab^2}{b^2+2}+\dfrac{bc^2}{c^2+2}+\dfrac{ca^2}{c^2+2}\right)\)
Ta có:
\(\dfrac{ab^2}{b^2+2}=\dfrac{2ab^2}{2b^2+4}=\dfrac{2ab^2}{b^2+b^2+4}\le\dfrac{2ab^2}{3\sqrt[3]{4b^4}}=\dfrac{a}{3}\sqrt[3]{2b^2}=\dfrac{a}{3}\sqrt[3]{2.b.b}\le\dfrac{a}{9}\left(2+b+b\right)\)
Tương tự và cộng lại:
\(VT\ge6-\left(\dfrac{2a}{9}\left(b+1\right)+\dfrac{2b}{9}\left(c+1\right)+\dfrac{2c}{9}\left(a+1\right)\right)\)
\(=6-\dfrac{2}{9}\left(a+b+c\right)-\dfrac{2}{9}\left(ab+bc+ca\right)\ge6-\dfrac{2}{9}\left(a+b+c\right)-\dfrac{2}{27}\left(a+b+c\right)^2=2\)
Dấu "=" xảy ra khi \(a=b=c=1\)
chứng minh \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\sqrt{\dfrac{2c}{a+b}}\ge2\) với mọi a,b,c >0
Ta có: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\sqrt{\dfrac{2c}{a+b}}\)
\(=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{2c}{\sqrt{2c\left(a+b\right)}}\)
\(\ge\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b+2c}=\dfrac{\left(a-b\right)^2\left(a+b+c\right)}{\left(b+c\right)\left(c+a\right)\left(a+b+2c\right)}\ge0\)
(đúng hiển nhiên)
Đẳng thức xảy ra khi $a=b=c.$
Cho \(a,b>0\); \(c< 0\). Chứng minh rằng:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
Cần cm:
\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\\ \Leftrightarrow a+b=a+b+2c+2\sqrt{\left(a+c\right)\left(b+c\right)}\\ \Leftrightarrow2c+2\sqrt{ab+ac+bc+c^2}=0\\ \Leftrightarrow2c+2\sqrt{c^2}=0\\ \Leftrightarrow2c+2\left|c\right|=0\\ \Leftrightarrow2c-2c=0\left(c< 0\right)\\ \Leftrightarrow0=0\left(luôn.đúng\right)\)
Vậy đẳng thức đc cm
Chứng minh a,b,c số thực không âm thỏa ab+bc+ca > 0 \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{2c+1}}+2\sqrt{\dfrac{c}{a+b+c}}\ge2\)
cho a,b,c>0 thỏa a+b+c=6
CMR \(\dfrac{a}{\sqrt{b^3+1}}+\dfrac{b}{\sqrt{c^3+1}}+\dfrac{c}{\sqrt{a^3+1}}\ge2\)
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
Cho \(a,b,c>0\). Chứng minh:
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}\)
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}=2\) (1)
\(VP=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{c}{\sqrt{c\left(a+b\right)}}\)
\(VP\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=2\) (2)
(1);(2) \(\Rightarrow VT< VP\)
Cho a,b,c > 0. Chứng minh:
a, a + b \(\ge2\sqrt{ab}\)
b, \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{ac}\)
a/ Xét hiệu: \(a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng) (đpcm)
''='' xảy ra khi a = b
b/ Sửa đề chút nhé: CMR:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ac}}\)
Áp dụng bđt AM-GM có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{a}\cdot\dfrac{1}{b}}=2\sqrt{\dfrac{1}{ab}}=\dfrac{2}{\sqrt{ab}}\);
Tương tự ta có:
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{\sqrt{bc}}\); \(\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{2}{\sqrt{ac}}\)
Cộng 2 vế ba bđt trên ta được:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge2\left(\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ac}}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ac}}\left(đpcm\right)\)
''='' xảy ra khi a = b = c
Cho a , b , c > 0. Chứng minh bất đẳng thức:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\sqrt{\dfrac{2c}{a+b}}\ge2\)