Cho a,b,c>0 thoã mãn: \(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}\ge1\)
Chứng minh rằng: (a+b)(b+c)(c+a) \(\le\)1
Cho a,b,c>0 thỏa mãn \(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}\ge1\). Chứng minh rằng \(a+b+c\ge ab+bc+ca\)
\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}\ge1\)
\(\Leftrightarrow2\ge\dfrac{a+b}{a+b+1}+\dfrac{b+c}{b+c+1}+\dfrac{c+a}{c+a+1}=\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+a+b}+\dfrac{\left(b+c\right)^2}{\left(b+c\right)^2+b+c}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)^2+c+a}\)
\(\Rightarrow2\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca+a+b+c}\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)+2\left(a+b+c\right)\ge2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)
\(\Rightarrow\)đpcm
cho a b c là 3 số dương thoã mãn a+b+c=1 chứng minh rằng:
\(\dfrac{c+ab}{a+b}\)+\(\dfrac{a+bc}{b+c}\)+\(\dfrac{b+ac}{a+c}\)≥2
Đặt vế trái là P
\(P=\dfrac{1.c+ab}{a+b}+\dfrac{1.a+bc}{b+c}+\dfrac{1.b+ac}{a+c}=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)
\(P=\dfrac{ac+c^2+bc+ab}{a+b}+\dfrac{a^2+ac+ab+bc}{b+c}+\dfrac{ab+ac+b^2+bc}{a+c}\)
\(P=\dfrac{c\left(a+c\right)+b\left(a+c\right)}{a+b}+\dfrac{a\left(a+c\right)+b\left(a+c\right)}{b+c}+\dfrac{a\left(b+c\right)+b\left(b+c\right)}{a+c}\)
\(P=\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)
Áp dụng BĐT Cô-si:
\(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}\ge2\sqrt{\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)}}=2\left(a+c\right)\) (1)
Tương tự: \(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(b+c\right)\) (2)
\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\) (3)
Cộng vế với vế (1);(2);(3):
\(2.\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+2.\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+2.\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+b\right)+2\left(b+c\right)+2\left(c+a\right)\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+c}\ge2\left(a+b+c\right)=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho các số thực a,b,c thỏa mãn điều kiện \(a\ge1,b\ge1,c\ge1\)
Chứng minh rằng : \(\dfrac{1}{2a-1}+\dfrac{1}{2b-1}+\dfrac{1}{2c-1}+\dfrac{4ab}{ab+1}+\dfrac{4bc}{bc+1}+\dfrac{4ac}{ac+1}\ge9\)
\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)
\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)
Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)
\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)
\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)
\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)
Cho a,b,c>0 thỏa mãn \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\ge1\). Chứng minh rằng:
a+b+c\(\ge\)ab+bc+ca
\(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\ge1\Leftrightarrow\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)
\(\Leftrightarrow\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}\le1\)
\(\Rightarrow1\ge\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\)
\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow\) đpcm
Cho 3 số thực dương a,b,c thoả mãn:\(abc\ge1\) .Chứng minh rằng :
\(\left(a+\dfrac{1}{a+1}\right)\left(b+\dfrac{1}{b+1}\right)\left(c+\dfrac{1}{c+1}\right)\ge\dfrac{27}{8}\)
\(a+\dfrac{1}{a+1}=\dfrac{a^2+a+1}{a+1}=\dfrac{4a^2+4a+4}{4\left(a+1\right)}=\dfrac{3\left(a+1\right)^2+\left(a-1\right)^2}{4\left(a+1\right)}\ge\dfrac{3\left(a+1\right)^2}{4\left(a+1\right)}=\dfrac{3}{4}\left(a+1\right)\ge\dfrac{3}{2}\sqrt{a}\)
Tương tự: \(b+\dfrac{1}{b+1}\ge\dfrac{3}{2}\sqrt{b}\) ; \(c+\dfrac{1}{c+1}\ge\dfrac{3}{2}\sqrt{c}\)
Nhân vế:
\(VT\ge\dfrac{27}{8}\sqrt{abc}\ge\dfrac{27}{8}\) (đpcm)
Cho a,b,c>0 thỏa mãn a+b+c=3 Cm\(\dfrac{1}{a^2+a+1}+\dfrac{1}{b^2+b+1}+\dfrac{1}{c^2+c+1}\ge1\)
\(\dfrac{1}{a^2+a+1}+\dfrac{1}{b^2+b+1}+\dfrac{1}{c^2+c+1}\ge1\)
\(\dfrac{1}{a^2+a+1}\ge\dfrac{1}{a^2+\dfrac{a^2+1}{2}+1}=\dfrac{2}{3}.\dfrac{1}{a^2+1}=\dfrac{2}{3}\left(1-\dfrac{a^2}{a^2+1}\right)\ge\dfrac{2}{3}\left(1-\dfrac{a}{2}\right)\)
Tương tự và cộng lại: \(VT\ge\dfrac{2}{3}\left(3-\dfrac{a+b+c}{2}\right)=\dfrac{2}{3}.\dfrac{3}{2}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a,b,c > 0 thoả mãn: a+b+c=1
chứng minh rằng: \(\dfrac{x}{1+y-x}\)+\(\dfrac{y}{1+z-y}\)+\(\dfrac{z}{1+x-z}\)\(\ge1\)
Chắc là a;b;c hết chứ?
\(VT=\dfrac{a}{a+b+c+b-a}+\dfrac{b}{a+b+c+c-b}+\dfrac{c}{a+b+c+a-c}\)
\(VT=\dfrac{a}{c+2b}+\dfrac{b}{a+2c}+\dfrac{c}{b+2a}=\dfrac{a^2}{ac+2ab}+\dfrac{b^2}{ab+2bc}+\dfrac{c^2}{bc+2ac}\)
\(VT\ge\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\) (đpcm)
cho x,y,z>0 ,x+y+z=1 chu nhi?
\(\Rightarrow\dfrac{x}{x+y+z+y-x}=\dfrac{x}{2y+z}\)
\(\Rightarrow\dfrac{y}{1+z-y}=\dfrac{y}{x+y+z+z-y}=\dfrac{y}{2z+x}\)
\(\Rightarrow\dfrac{z}{1+x-z}=\dfrac{z}{x+y+z+x-z}=\dfrac{z}{2x+y}\)
\(\Rightarrow A=\dfrac{x}{2y+z}+\dfrac{y}{2z+x}+\dfrac{z}{2x+y}=\dfrac{x^2}{2xy+xz}+\dfrac{y^2}{2zy+xy}+\dfrac{z^2}{2xz+xz}\ge\dfrac{\left(x+y+z\right)^2}{3\left(xy+yz+xz\right)}=1\)
dau"=" xay ra<=>x=y=z=1/3
Cho a,b,c>0 thỏa mãn a+b+c=\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\). Chứng minh rằng:
\(\dfrac{1}{a^3+b+c}+\dfrac{1}{a+b^3+c}+\dfrac{1}{a+b+c^3}\le1\)
cho a,b,c\(\ge\)0,a+b+c=1.chứng minh rằng
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{9}{10}\)
Ta có đánh giá sau với a không âm:
\(\dfrac{a}{1+a^2}\le\dfrac{36a+3}{50}\)
Thật vậy, BĐT tương đương:
\(\left(36a+3\right)\left(a^2+1\right)\ge50a\)
\(\Leftrightarrow\left(3a-1\right)^2\left(4a+3\right)\ge0\) (luôn đúng)
Tương tự: \(\dfrac{b}{1+b^2}\le\dfrac{36b+3}{50}\) ; \(\dfrac{c}{1+c^2}\le\dfrac{36c+3}{50}\)
Cộng vế: \(VT\le\dfrac{36\left(a+b+c\right)+9}{50}=\dfrac{9}{10}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)
Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)
\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)
\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)
Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)
Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)