Cho a,b,c dương. C/m\(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\ge\frac{a+b+c}{2}\)
Cho a,b,c dương và a+b+c=3. c/m\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{2}\)
và \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\ge\frac{3}{2}\)
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{2}\)
+) cm: \(\frac{1}{a^2+1}=1-\frac{a^2}{a^2+1}\ge1-\frac{a^2}{2a}=1-\frac{a}{2}\)
\(\frac{1}{b^2+1}\ge1-\frac{b}{2}\)
\(\frac{1}{c^2+1}\ge1-\frac{c}{2}\)
Cộng theo vế:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge3-\frac{a+b+c}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
Cho a, b, c, d dương. CM:
1) \(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
2) \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b+c}{\sqrt[3]{abc}}\)
3) \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{d^2}+\frac{d^2}{a^2}\ge\frac{a+b+c+d}{\sqrt[4]{abcd}}\)
4) \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9;a+b+c\le1\)
Làm tạm một câu rồi đi chơi, lát làm cho.
4)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
2/ Cô: \(\frac{2a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a.a.b}{b.b.c}}=3\sqrt[3]{\frac{a^3}{abc}}=\frac{3a}{\sqrt[3]{abc}}\)
Tương tự hai BĐT còn lại và cộng theo vế thu được:
\(3.VT\ge3.VP\Rightarrow VT\ge VP^{\left(Đpcm\right)}\)
Đẳng thức xảy ra khi a = b= c
Cho a,b,c là 3 số dương t/m: a+b+c=3
CMR:\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\)
#)Giải :
Áp dụng BĐT Cauchy : \(\hept{\begin{cases}\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}\\\frac{b}{1+c^2}=b-\frac{bc^2}{1+c^2}\ge b-\frac{bc}{2}\\\frac{c}{1+a^2}=c-\frac{ca^2}{1+a^2}\ge c-\frac{ca}{2}\end{cases}}\)
\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{1}{2}\left(ab+bc+ca\right)\ge3-\frac{1}{6}\left(a+b+c\right)^2=\frac{3}{2}\)
\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\left(đpcm\right)\)
Theo BĐT AM-GM:
\(\frac{a}{1+b^2}\)=a-\(\frac{ab^2}{1+b^2}\)\(\ge\)a-\(\frac{ab^2}{2b}\)=a-\(\frac{ab}{2}\)
Tương tự: \(\frac{b}{1+c^2}\)\(\ge\)b-\(\frac{bc}{2}\);\(\frac{c}{1+a^2}\)\(\ge\)c-\(\frac{ca}{2}\)
Suy ra \(\frac{a}{1+b^2}\)+\(\frac{b}{1+c^2}\)+\(\frac{c}{1+a^2}\)\(\ge\)a+b+c-\(\frac{1}{2}\)(ab+bc+ca)
Mặt khác thì theo BĐT AM-GM:9=a2+b2+c2+2(ab+bc+ca)
=\(\frac{a^2+b^2}{2}\)+\(\frac{b^2+c^2}{2}\)+\(\frac{c^2+a^2}{2}\)+2(ab+bc+ca)\(\ge\)3(ab+bc+ca)
\(\Rightarrow\)\(\frac{1}{2}\)(ab+bc+ca)\(\le\)\(\frac{3}{2}\)
Cho nên \(\frac{a}{1+b^2}\)+\(\frac{b}{1+c^2}\)+\(\frac{c}{1+a^2}\)\(\ge\)a+b+c-\(\frac{3}{2}\)=3-\(\frac{3}{2}\)=\(\frac{3}{2}\)
cho a,b,c dương. cmr
a, \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
b, \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
Câu a : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\ge\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{9}{2}\)
\(VT=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{\left(a+b+c\right).9}{2\left(a+b+c\right)}=\frac{9}{2}\) (đpcm)
Dấu "\("="\) xảy ra khi \(a=b=c\)
Câu b : \(VT=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\left(đpcm\right)\)
Dấu = xảy ra khi a=b=c
Cho a, b, c là các số thực dương và thỏa mãn: a + b + c = 3.
CMR: \(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\ge\frac{3}{2}\)
\(P=\sum\frac{a^2}{a+b^2}=\sum\left(a-\frac{ab^2}{a+b^2}\right)\ge\sum\left(a-\frac{ab^2}{2b\sqrt{a}}\right)=\sum\left(a-\frac{1}{2}b\sqrt{a}\right)\)
\(P\ge\sum\left(a-\frac{1}{2}\sqrt{b}.\sqrt{ab}\right)\ge\sum\left(a-\frac{1}{4}\left(b+ab\right)\right)\)
\(P\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(ab+bc+ca\right)\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{12}\left(a+b+c\right)^2=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho các số thực dương a,b,c. CMR
\(\frac{a^3}{a^2+b^2}+\frac{b^3}{c^2+b^2}+\frac{c^3}{a^2+c^2}\ge\frac{a+b+c}{2}\)
Lời giải:
\(\text{VT}=a-\frac{ab^2}{a^2+b^2}+b-\frac{bc^2}{b^2+c^2}+c-\frac{ca^2}{a^2+c^2}=(a+b+c)-\left(\frac{ab^2}{a^2+b^2}+\frac{bc^2}{b^2+c^2}+\frac{ca^2}{c^2+a^2}\right)(1)\)
Áp dụng BĐT AM-GM:
\(\frac{ab^2}{a^2+b^2}+\frac{bc^2}{b^2+c^2}+\frac{ca^2}{c^2+a^2}\leq \frac{ab^2}{2ab}+\frac{bc^2}{2bc}+\frac{ca^2}{2ac}=\frac{a+b+c}{2}(2)\)
Từ $(1);(2)\Rightarrow \text{VT}\geq a+b+c-\frac{a+b+c}{2}=\frac{a+b+c}{2}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Cho các số dương a,b,c.CMR :\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}\ge\frac{a+b+c}{3}\)
Lời giải:
Áp dụng BĐT AM-GM: \(ab\leq \frac{a^2+b^2}{2}\Rightarrow a^2+ab+b^2\leq \frac{3}{2}(a^2+b^2)\)
\(\Rightarrow \frac{a^3}{a^2+ab+b^2}\geq \frac{2}{3}.\frac{a^3}{a^2+b^2}=\frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\)
Mà cũng theo BĐT AM-GM: \(\frac{ab^2}{a^2+b^2}\leq \frac{ab^2}{2ab}=\frac{b}{2}\)
\(\Rightarrow \frac{a^3}{a^2+ab+b^2}\geq \frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\geq \frac{2}{3}(a-\frac{b}{2})\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\Rightarrow \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}\geq \frac{2}{3}(a-\frac{b}{2})+\frac{2}{3}(b-\frac{c}{2})+\frac{2}{3}(c-\frac{a}{2})=\frac{a+b+c}{3}\)
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Ta có:\(\frac{a^3}{a^2+ab+b^2}=\frac{a\left(a^2+ab+b^2\right)-ab\left(a+b\right)}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\)
Lại có:\(a^2+ab+b^2\ge3ab\)
\(\Rightarrow a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3ab}=a-\frac{a+b}{3}\)
\(\Rightarrow\sum\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
"="<=>a=b=c
Cho 3 số dương a,b,c thỏa mãn a+b+c=3. CMR
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\)
\(\frac{a}{1+b^2}=\frac{a\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự ta có \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\); \(\frac{c}{1+a^2}\ge c-\frac{ac}{2}\)
\(\Rightarrow VT\ge a+b+c-\frac{1}{2}\left(ab+ac+bc\right)\ge3-\frac{1}{6}\left(a+b+c\right)^2=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho các số a,b,c dương. Chứng minh rằng \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2\sqrt{a}}{a^3+b^2}+\frac{2\sqrt{b}}{b^3+c^2}+\frac{2\sqrt{c}}{c^3+a^2}\)
Ta có: Theo bất đẳng thức cauchy schwarz và bất đẳng thức cauchy với a;b;c>0 ta có:
\(\dfrac{1}{a^2}+\dfrac{1}{a^2}=\dfrac{\left(\sqrt{a}\right)^2}{a^3}+\dfrac{1}{a^2}\ge\dfrac{\left(\sqrt{a}+1\right)^2}{a^3+a^2}\ge\dfrac{4\sqrt{a}}{a^3+a^2}\)(1)
Tương tự \(\dfrac{1}{b^2}+\dfrac{1}{b^2}\ge\dfrac{4\sqrt{b}}{b^3+b^2}\left(2\right);\dfrac{1}{c^2}+\dfrac{1}{c^2}\ge\dfrac{4\sqrt{c}}{c^3+c^2}\left(3\right)\)
Cộng từng vế (1) ;(2);(3) vế theo vế rồi chia hai vế cho 2 ta có đpcm
Áp dụng BĐT Cauchy schwarz kết hợp với AM-GM cho các số dương ta có :
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{a}{a^3}+\dfrac{1}{b^2}\ge\dfrac{\left(\sqrt{a}+1\right)^2}{a^3+b^2}\ge\dfrac{4\sqrt{a}}{a^3+b^2}\)
\(\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{b}{b^3}+\dfrac{1}{c^2}\ge\dfrac{\left(\sqrt{b}+1\right)^2}{b^3+c^2}\ge\dfrac{4\sqrt{b}}{b^3+c^2}\)
\(\dfrac{1}{c^2}+\dfrac{1}{a^2}=\dfrac{c}{c^3}+\dfrac{1}{a^2}\ge\dfrac{\left(\sqrt{c}+1\right)^2}{c^3+a^2}\ge\dfrac{4\sqrt{c}}{c^3+a^2}\)
Cộng từng vế của BĐT ta được :
\(2\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{4\sqrt{a}}{a^3+b^2}+\dfrac{4\sqrt{b}}{b^3+c^2}+\dfrac{4\sqrt{c}}{c^3+a^2}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2\sqrt{a}}{a^3+b^2}+\dfrac{2\sqrt{b}}{b^3+c^2}+\dfrac{2\sqrt{c}}{c^3+a^2}\) ( đpcm )
Dấu \("="\) xảy ra khi \(a=b=c\)