Cho a, b, c > 0 và \(a+b+c\le3\) . CMR :
A= \(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{3}{2}\)
Cho a,b,c>0 thỏa mãn a+b+c\(\le3\).
CMR : A= \(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{3}{2}\)
(Sử dụng Cauchy)
Đầu tiên ta cm bđt:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)(tự cm)
Áp dụng ta có:
\(A=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{9}{3+ab+bc+ca}\)
Cần cm:\(ab+bc+ca\le3\)
Hay \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
=>đpcm
cho a,b,c>0 thỏa mãn abc=1.
CMR:\(\dfrac{a}{ab+1}+\dfrac{b}{bc+1}+\dfrac{c}{ca+1}\ge\dfrac{3}{2}\)
Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
Cho các số thực dương a,b và c thoả mãn: \(\dfrac{1}{a+2}\)+\(\dfrac{1}{b+2}\)+\(\dfrac{1}{c+2}\)\(\ge\dfrac{3}{2}\)
CMR: \(a+b+c\ge ab+bc+ca\)
\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)
\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)
\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)
\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)
\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow\) đpcm
Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ
Cho a,b,c>0 thỏa mãn\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1\). CMR
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\)
Áp dụng BĐT BSC:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)
\(=\dfrac{a+b+c}{2}\)
\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
cho a,b,c>0;\(a+b+c,abc=1\).CMR
\(\dfrac{bc}{a^2\left(b+c\right)}+\dfrac{ca}{b^2\left(c+a\right)}+\dfrac{ab}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
cho 3 số a,b,c dương và a+b+c=1.CMR
\(\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{15}{4}\)
Cho a,b,c>0 thỏa mãn a+b+c\(\le\)3
CMR:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{3}{2}\)
(Sử dụng Cauchy)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(A=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)
\(A\ge\dfrac{\left(1+1+1\right)^2}{3+ab+bc+ac}=\dfrac{9}{3+ab+bc+ac}\)
Mặt khác,theo hệ quả AM-GM: \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}\le\dfrac{3^2}{3}=3\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ac}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Cho các số thực dương thoả mãn:
\(\dfrac{1}{a+2}\) + \(\dfrac{1}{b+2}\)+ \(\dfrac{1}{c+2}\)\(\ge1\). CMR: \(a+b+c\ge ab+bc+ca\)
Cho các số dương a,b,c thỏa mãn ab + bc + ca = 3. CMR:
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\)
Lời giải:
Không mất tổng quát giả sử \(c=\min (a,b,c)\)
Khi đó, do \(ab+bc+ac=3\Rightarrow ab\geq 1\).
Với $ab\geq 1$ ta có bổ đề sau: \(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\)
Để cm bổ đề trên rất đơn giản. Quy đồng và biến đổi tương đương thu được \((a-b)^2(ab-1)\geq 0\) (luôn đúng với mọi \(ab\geq 1\) )
Sử dụng bổ đề vào bài toán:
\(\Rightarrow \text{VT}\geq \frac{2}{ab+1}+\frac{1}{c^2+1}=\frac{2c^2+ab+3}{abc^2+ab+c^2+1}(*)\)
Giờ ta sẽ cm \(\frac{2c^2+ab+3}{abc^2+ab+c^2+1}\geq \frac{3}{2}(**)\)
\(\Leftrightarrow 2(2c^2+ab+3)\geq 3(abc^2+ab+c^2+1)\)
\(\Leftrightarrow c^2+3\geq 3abc^2+ab\)
\(\Leftrightarrow c^2+bc+ac\geq 3abc^2\)
\(\Leftrightarrow c+b+a\geq 3abc\).
BĐT trên đúng do theo AM-GM: \(3(a+b+c)=(ab+bc+ac)(a+b+c)\geq 9abc\Rightarrow a+b+c\geq 3abc\) )
Do đó $(*)$ được cm.
Từ \((*),(**)\Rightarrow \text{VT}\geq \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=1$