Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
cho a,b,c ≥0.CMR
a+b+c ≥\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Cho các số thực dương a, b, c thỏa mãn: abc + a + b = 3ab. Chứng minh rằng:\(\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{b}{bc+b+1}}+\sqrt{\frac{a}{ca+c+1}}\ge\sqrt{3}\)
11 tháng 5 2017 lúc 17:01
Cho \(a,b,c\) là các số thực không âm. CMR:
\(3\left(a^2+b^2+c^2\right)\ge\) \(\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\) \(+\left(a-b\right)^2\) \(+\left(b-c\right)^2+\left(c-a\right)^2\ge\left(a+b+c\right)^2\)
19 tháng 6 2021 lúc 20:44
Cho các số thực dương a,b,c thảo mãn \(a^2+b^2+c^2=1\). CHứng minh:
\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}+\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}+\sqrt{\dfrac{ca+2b^2}{1+ca-b^2}}\ge2+ab+bc+ac\)
Cho a,b,c,d,e inR . Chứng minh các BĐT sau:
a/ a2 + b2 + c2 ge ab + bc + ca
b/ a2 + b2 +1 ge ab + a + b
c/ a2 + b2 +c2 + 3 ge 2( a + b + c)
d/ a2 + b2 + c2 ge 2( ab + bc - ca)
e/ a4 + b4 + c2 +1 ge 2a( ab2 - a +c +1)
f/ dfrac{a^2}{4}+ b2 + c2 ge ab - ac +2bc
g/ a2 (1+b2) + b2 (1+c2) +c2 (1+a2) ge 6abc
h/ a2 +b2+ c2+ d2+ e2 ge a(b+c+d+e)
i/ dfrac{1}{a}+ dfrac{1}{b}+dfrac{1}{c} ge dfrac{1}{sqrt{ab}}+dfrac{1}{sqrt{bc}}+dfrac{1}{sqrt{ca}} , (a,b,c 0)
j/ a+b+c ge sqrt{ab}+sqrt{bc}+sqrt{ca...
Đọc tiếp
Cho a,b,c,d,e \(\in\)\(R\) . Chứng minh các BĐT sau:
a/ a2 + b2 + c2 \(\ge\) ab + bc + ca
b/ a2 + b2 +1 \(\ge\) ab + a + b
c/ a2 + b2 +c2 + 3 \(\ge\) 2( a + b + c)
d/ a2 + b2 + c2 \(\ge\) 2( ab + bc - ca)
e/ a4 + b4 + c2 +1 \(\ge\) 2a( ab2 - a +c +1)
f/ \(\dfrac{a^2}{4}\)+ b2 + c2 \(\ge\) ab - ac +2bc
g/ a2 (1+b2) + b2 (1+c2) +c2 (1+a2) \(\ge\) 6abc
h/ a2 +b2+ c2+ d2+ e2 \(\ge\) a(b+c+d+e)
i/ \(\dfrac{1}{a}\)+ \(\dfrac{1}{b}\)+\(\dfrac{1}{c}\) \(\ge\) \(\dfrac{1}{\sqrt{ab}}\)+\(\dfrac{1}{\sqrt{bc}}\)+\(\dfrac{1}{\sqrt{ca}}\) , (a,b,c > 0)
j/ a+b+c \(\ge\) \(\sqrt{ab}\)+\(\sqrt{bc}\)+\(\sqrt{ca}\) ( a,b,c \(\ge\)0)
Rút gọn các biểu thức
a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0)
b) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0; a ≠ b)
c) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)\) với a,b > 0
\(a+b+c\ge\sqrt{ab}+\sqrt{bc}\)
Bài 1: CMR:
a, (4+sqrt{3}). (4-sqrt{3})13
b, sqrt{8+2sqrt{7}}-sqrt{8-2sqrt{7}}2
c, frac{sqrt{1}}{2+sqrt{3}}+frac{sqrt{1}}{2-sqrt{3}}4
d, frac{asqrt{b}+bsqrt{a}}{sqrt{ab}}:frac{1}{sqrt{a}-sqrt{b}}a-b(a0, b0, a≠b)
Bài 2: CMR:
a, sqrt{a}+frac{sqrt{1}}{sqrt{a}}ge2left(a0right)
b, a+b+frac{1}{2}gesqrt{a}+sqrt{b}left(a,b0right)
c, frac{1}{x}+frac{1}{y}+frac{1}{z}gefrac{1}{sqrt{xyz}}+frac{1}{sqrt{yz}}+frac{1}{sqrt{zx}}left(x,y,z0right)
d, frac{sqrt{3}+2}{sqrt{3}-2}-frac{sqrt{3}-2}{sqrt{3}+2}-8...
Đọc tiếp
Bài 1: CMR:
a, (4+\(\sqrt{3}\)). (4-\(\sqrt{3}\))=13
b, \(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=2\)
c, \(\frac{\sqrt{1}}{2+\sqrt{3}}+\frac{\sqrt{1}}{2-\sqrt{3}}=4\)
d, \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=a-b\)(a>0, b>0, a≠b)
Bài 2: CMR:
a, \(\sqrt{a}+\frac{\sqrt{1}}{\sqrt{a}}\ge2\left(a>0\right)\)
b, a+b+\(\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\left(a,b>0\right)\)
c, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xyz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\left(x,y,z>0\right)\)
d, \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=-8\sqrt{3}\)
e, \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)=a-b(a>0, b>0, a≠b)
Bài 3: Tìm Min hoặc Max(nếu có):
a, \(\sqrt{x^2+9}\)
b, \(\frac{2}{\sqrt{x^2+1}}\)
c, 1-\(\sqrt{5+2x-x^2}\)
cho a,b,c ≥0. CMR:
a+b+\(\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)