Cho a,b,c >0 tm abc=1, C/m
\(\dfrac{1}{\sqrt{a^5+b^2+ab+6}}+\dfrac{1}{\sqrt{b^5+c^2+bc+6}}+\dfrac{1}{\sqrt{c^5+a^2+ca+6}}\le1\)
Cho a,b,c>0 t/m \(a^2+b^2+c^2=1\).
C/m \(\dfrac{1}{4-\sqrt{ab}}+\dfrac{1}{4-\sqrt{bc}}+\dfrac{1}{4-\sqrt{ca}}\le1\)
Đề bài sai, bạn kiểm tra lại điều kiện \(a^2+b^2+c^2=1\)
Bài 1:
a , Cho a , b là các số dương . C/m: \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab}\)
b, Cho a , b , c là các số dương thoả mãn a+b+c+ab+bc+ca=6abc
C/m: \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)
Bài 2:a, Cho a, b ,c là các số thực không âm thỏa mãn a+b+c=1
C/m: \(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\)
b,C/m: \(\dfrac{a+b+c}{\sqrt{a\left(a+3b\right)}+\sqrt{b\left(b+2c\right)}+\sqrt{c\left(c+2a\right)}}\ge\dfrac{1}{2}\)
Bài 3: Cho a , b, c> 0 thỏa mãn abc=1. Tìm max của:
\(P=\dfrac{ab}{a^5+b^5+ab}+\dfrac{bc}{b^5+c^5+bc}+\dfrac{ca}{c^5+a^5+ca}\)
1) Áp dụng bđt Cauchy:
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge2\sqrt{\dfrac{1}{a^2b^2}}=\dfrac{2}{ab}\)
Xong
Cho a, b, c thỏa mãn: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\). Tính giá trị của biểu thức: \(H=\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Cho a, b, c thoả mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\). Tính giá trị biểu thức: \(N=\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Cho a, b, c>0 thỏa mãn: abc=1. CM: \(\dfrac{1}{\sqrt{ab+a+2}}+\dfrac{1}{\sqrt{bc+b+2}}+\dfrac{1}{\sqrt{ca+c+2}}\le\dfrac{3}{2}\)
cho a,b,c>0 thỏa mãn \(a^2+b^2+c^2=1\).CMR
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}+\dfrac{\sqrt{bc+2a^2}}{\sqrt{1+bc-a^2}}+\dfrac{\sqrt{ca+2b^2}}{\sqrt{1+ca-b^2}}\ge2+ab+bc+ca\)
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
Cho a,b,c >0 thỏa mãn abc=1. Chứng minh:
\(\dfrac{1}{\sqrt{ab+a+2}}+\dfrac{1}{\sqrt{bc+b+2}}+\dfrac{1}{\sqrt{ca+c+2}}\le\dfrac{3}{2}\)
Lời giải:
Vì $abc=1$ nên tồn tại $x,y,z$ sao cho : \((a,b,c)=\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right)\)
Khi đó:
\(\text{VT}=\frac{1}{\sqrt{\frac{x}{z}+\frac{x}{y}+2}}+\frac{1}{\sqrt{\frac{y}{x}+\frac{y}{z}+2}}+\frac{1}{\sqrt{\frac{z}{y}+\frac{z}{x}+2}}=\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}+\frac{\sqrt{xz}}{\sqrt{xy+yz+2xz}}+\frac{\sqrt{xy}}{\sqrt{xz+yz+2xy}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}^2\leq (1+1+1)\left(\frac{yz}{xy+xz+2yz}+\frac{xz}{xy+yz+2xz}+\frac{xy}{xz+yz+2xy}\right)\)
\(\leq 3\left[\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)+\frac{xz}{4}\left(\frac{1}{xy+xz}+\frac{1}{xz+yz}\right)+\frac{xy}{4}\left(\frac{1}{xz+xy}+\frac{1}{yz+xy}\right)\right]\)
hay \(\text{VT}^2\leq \frac{3}{4}.\left(\frac{xy+yz}{xy+yz}+\frac{xy+xz}{xy+xz}+\frac{yz+xz}{yz+xz}\right)=\frac{9}{4}\)
\(\Rightarrow \text{VT}\leq \frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$
cho a,b,c là các số thực dương thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge1\)
chứng minh rằng \(\dfrac{a+b}{\sqrt{ab+c}}+\dfrac{b+c}{\sqrt{bc+a}}+\dfrac{c+a}{\sqrt{ca+b}}\ge3\sqrt[6]{abc}\)
Rút gọn: ( 2,5 Điểm )
A= \(\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}\)+ \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)
B= \(\dfrac{3}{\sqrt{5}-2}\)+ \(\dfrac{4}{\sqrt{6}+\sqrt{2}}\)+ \(\dfrac{1}{\sqrt{6}+\sqrt{5}}\)
C = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
D= \(\dfrac{1}{2-\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
E = \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
F = \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)