cho ΔABC có \(\widehat{A}=60^o\). đaẹt BC=a; CA=b; AB=c. CMR: \(\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}\)
cho a,b,c>0 thỏa mãn a+b+c=1. CMR: \(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{3}{2}\)
Cho tan giác ABC có: \(\widehat{C}=2\widehat{B}=4\widehat{A}\). CMR: \(\dfrac{1}{AB}+\dfrac{1}{AC}=\dfrac{1}{BC}\)
Cho tam giác ABC có \(\widehat{C}=2\widehat{B}=4\widehat{A}\). CMR: \(\dfrac{1}{AB}+\dfrac{1}{AC}=\dfrac{1}{BC}\)
cho a,b,c>0.CMR
\(\dfrac{a+b}{ab+c^2}+\dfrac{b+c}{bc+a^2}+\dfrac{c+a}{ca+b^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\dfrac{a+b}{ab+c^2}=\dfrac{\left(a+b\right)^2}{\left(ab+c^2\right)\left(a+b\right)}=\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\le\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\)
Tương tự:
\(\dfrac{b+c}{bc+a^2}\le\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}\) ; \(\dfrac{c+a}{ca+b^2}\le\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)
Cộng vế:
\(VT\le\dfrac{1}{a}\left(\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)+\dfrac{1}{b}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{c^2}{a^2+c^2}\right)+\dfrac{1}{c}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Cho 3 số a,b,c thỏa mãn ab + bc + ca = 1. CMR:
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
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 a,b,c >0 thỏa \(a^2+b^2+c^2=1.CMR:\)
\(P=\dfrac{bc}{a^2+1}+\dfrac{ca}{b^2+1}+\dfrac{ab}{c^2+1}\le\dfrac{3}{4}\)
Lời giải:Áp dụng BĐT AM-GM và BĐT Cauchy-Schwarz:
\(\frac{bc}{a^2+1}=\frac{bc}{(a^2+b^2)+(a^2+c^2)}\leq \frac{1}{4}.\frac{(b+c)^2}{(a^2+b^2)+(a^2+c^2)}\leq \frac{1}{4}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)\)
Hoàn toàn tương tự với các phân thức còn lại, ta có:
\(P\leq \frac{1}{4}\left(\frac{b^2+a^2}{a^2+b^2}+\frac{c^2+a^2}{a^2+c^2}+\frac{b^2+c^2}{b^2+c^2}\right)=\frac{3}{4}\)
(đpcm)
Dấu "=" xảy ra khi $a=b=c=\sqrt{\frac{1}{3}}$
Cho a, b, c > 0 và ab + bc + ca = 1.
CMR : \(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}>\dfrac{1}{2}\)
Ta có :
\(VT=\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)
Theo BĐT Cauchy ta có :
\(\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\)
Theo BĐT Cô - Si ta lại có : \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Rightarrow VT\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}=\dfrac{1}{2}\)
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