Chứng minh rằng
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
Cho a,b,c >0 Chứng minh rằng:
a) \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)
b) \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
Chứng minh rằng với mọi a, b, c ta có: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{a^2}{c^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{a}{c}\)
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 \(a=b=c=\dfrac{1}{3}\)
Cho ba số thực dương a, b, c . Chứng minh rằng:
\(\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a++b}\ge a+b+c\)
\(\dfrac{a^2+bc}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)-a\left(b+c\right)}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}-a\)
\(\Rightarrow VT=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}-\left(a+b+c\right)\)
Mặt khác áp dụng \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Rightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge a+b+b+c+a+c=2\left(a+b+c\right)\)
\(\Rightarrow VT\ge2\left(a+b+c\right)-\left(a+b+c\right)=a+b+c\) (đpcm)
Cho a;b;c>0 Chứng minh rằng: \(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge\dfrac{a+b+c}{2}\)
\(\sum\dfrac{a^3}{a^2+b^2}=a+b+c-\dfrac{ab^2}{a^2+b^2}-\dfrac{bc^2}{b^2+c^2}-\dfrac{ca^2}{c^2+a^2}\ge a+b+c-\dfrac{b}{2}-\dfrac{c}{2}-\dfrac{a}{2}=\dfrac{a+b+c}{2}\) Dấu "=" xảy ra khi: \(a=b=c\)
Cho a,b,c >0. Chứng minh rằng:
\(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)
\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
giả sử \(a>b>c>0\) thì ta có :
\(\dfrac{a^2}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^2}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\ge2\dfrac{a}{b}+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\)
\(=\dfrac{2a}{b}+\dfrac{c^3}{a^3}-\dfrac{c^2}{a^2}\ge0\)
làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
Cho a,b,c là các số dương. Chứng minh rằng:\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{b+a}\ge\dfrac{a+b+c}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{b+c+a+c+a+b}\)
\(=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=VP\)
Cho a,b,c>0.Chứng minh rằng:
\(\dfrac{a+b}{a^2+b^2}+\dfrac{b+c}{b^2+c^2}+\dfrac{a+c}{a^2+c^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Help me?!
Cho a,b,c dương. Chứng minh rằng:
\(\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{c+a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}\ge\dfrac{15}{2}\)
\(A=\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{c+a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}\)
\(A=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\right)\)
\(A\ge\dfrac{3}{2}+\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\) (bất đẳng thức Nesbit)
\(A\ge\dfrac{3}{2}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}\)
\(A\ge\dfrac{3}{2}+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương ta có:
\(A\ge\dfrac{3}{2}+2\sqrt{\dfrac{ab}{ab}}+2\sqrt{\dfrac{ac}{ac}}+2\sqrt{\dfrac{bc}{bc}}\)
\(A\ge\dfrac{3}{2}+2+2+2=\dfrac{15}{2}\left(đpcm\right)\)
Dấu"=" xảy ra khi: \(a=b=c\)