chứng minh BĐT
\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\ge\dfrac{9}{2}\)
Cho a, b, c > . Chứng minh rằng:
a, \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
b, \(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
a.
Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
b.
Ta có:
\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)
Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
Cộng vế với vế:
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
1: Cho a,b,c>0. CMR: \(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ac}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{c^2+a^2}{b^2+ac}\ge\dfrac{9}{2}\)
Áp dụng BĐT AM - GM, ta có:
\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\)
\(\ge\dfrac{3\sqrt{a^3b^3c^3}}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{a^2+c^2}{b^2+\dfrac{a^2+c^2}{2}}\)
\(\ge\dfrac{3abc}{2abc}+\dfrac{2\left(a^2+b^2\right)}{2c^2+a^2+b^2}+\dfrac{2\left(b^2+c^2\right)}{2a^2+b^2+c^2}+\dfrac{2\left(a^2+c^2\right)}{2b^2+a^2+c^2}\)
\(=\dfrac{3}{2}+2\times\left[\dfrac{a^2+b^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\dfrac{b^2+c^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\dfrac{c^2+a^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\right]\) (1)
Đặt \(\left\{{}\begin{matrix}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{matrix}\right.\), ta có:
\(\left(1\right)\Leftrightarrow\dfrac{3}{2}+2\times\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\)
\(\ge\dfrac{3}{2}+2\times\dfrac{3}{2}\) (Bất_đẳng_thức_Nesbitt)
\(=\dfrac{9}{2}\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c
cho a,b,c >0 . Chứng minh
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{\left(a+b+c\right)}{3}\)
AM-GM ngược dấu như sau:
\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3ab}=\dfrac{2a-b}{3}\)
Tương tự ta cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{2b-c}{3};\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{2c-a}{3}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a-b}{3}+\dfrac{2b-c}{3}+\dfrac{2c-a}{3}=\dfrac{a+b+c}{3}=VP\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)
Dễ thấy :
\(a^{3}+b^{3}+c^{3}+ab(b+c)+bc(b+c)+ca(c+a)=(a^{2}+ b^{2}+c^{2})(a+b+c)\)
\(\Rightarrow VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
Vậy cần chứng minh
\(\dfrac{a^2+b^2+c^2}{a+b+c}\ge\dfrac{a+b+c}{3}\Leftrightarrow\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2\right)\) (luôn đúng)
CMR: Với các số thực dương a;b;c thì\(\dfrac{a^3+2abc+b^3}{c^2+ab}+\dfrac{a^3+2abc+c^3}{b^2+ac}+\dfrac{b^3+2abc+c^3}{a^2+bc}\ge2\left(a+b+c\right)\)
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) \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) , Chứng Minh BĐT
Cho \(a,b,c\) là các số dương. Chứng minh: \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\ge\dfrac{a+b+c}{3}\)
a ) Cho a,b,c >0 C/m:
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{a^2+b^2+c^2}{a+b+c}\)
b ) Cho a,b,c > 0 . C/m :
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.\)
c ) Cho a,b,c > 0 . C/m :
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c.\)
giúp nha mn
a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
b)
Áp dụng BĐT Cauchy Shwarz, ta có:
\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2\)
Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)^3}{9}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)}{3}\times\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}\) (đpcm)
Dấu "=" xảy ra khi a = b = c.
Giải giùm mình mấy bài BPT này nha
a) Chứng minh: \(\dfrac{a+b}{2}\le\sqrt{\dfrac{a^2+b^2}{2}}\)
b) Cho a,b>0 chứng minh: \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)
c) Cho a+b\(\ge\)0 chứng minh: \(\dfrac{a+b}{2}\ge\sqrt[3]{\dfrac{a^3+b^3}{2}}\)
d) Chứng minh: \(\dfrac{a+b+c}{3}\ge\sqrt{\dfrac{ab+bc+ac}{3}}\) ; \(a,b,c\ge0\)
e) Chứng minh: \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM