Cho a + b + c = 3. Chứng minh \(\dfrac{a^2-bc}{a^2+3}+\dfrac{b^2-ca}{b^2+3}+\dfrac{c^2-ab}{c^2+3}\ge0\)
Cho \(a,b,c\) là các số dương. Chứng minh: \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\ge\dfrac{a+b+c}{3}\)
cho a,b,c >0 thỏa mãn \(a^2+b^2+c^2=3\) chứng minh rằng \(\dfrac{a}{ab+3}+\dfrac{b}{bc+3}+\dfrac{c}{ca+3}\le\dfrac{3}{4}\)
Chứng minh :
a, \(\dfrac{a+b+c}{3}\dfrac{>}{ }\sqrt{\dfrac{ab+bc+ca}{3}}\) với a,b,c>0
b,\(\dfrac{a^2+b^2+c^2}{3}\dfrac{>}{ }\left(\dfrac{a+b+c}{3}\right)^2\)
c,\(\dfrac{x^2+2}{\sqrt{x^2+1}}\dfrac{>}{ }2\)
d,\(\dfrac{a^3+b^3}{2}\dfrac{>}{ }\left(\dfrac{a+b}{2}\right)^3\)
a) Ta có:
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{9}\ge\dfrac{\left(ab+bc+ca\right)}{3}\)
\(\Leftrightarrow\dfrac{a+b+c}{3}\ge\sqrt{\dfrac{ab+bc+ca}{3}}\)
Đẳng thức xảy ra khi $a=b=c.$
b) BĐT \(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
Hay là \(2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\),
đúng.
Đẳng thức xảy ra khi $a=b=c.$
c) \(\Leftrightarrow\dfrac{\left(x^2+2\right)^2}{x^2+1}\ge4\Leftrightarrow x^4+4x^2+4\ge4x^2+4\Leftrightarrow x^4\ge0\)
Đẳng thức xảy ra khi $x=0.$
d) Xét hiệu hai vế đi bạn.
Chứng minh:
a, \(a^3+b^3+c^3\dfrac{>}{ }3abc\)
b,\(abc\dfrac{< }{ }\left(\dfrac{a+b+c}{3}\right)^3\)
c,\(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\dfrac{< }{ }a+b+c\)
d,\(\dfrac{a}{b+c}+\dfrac{c}{a+b}+\dfrac{b}{a+c}\dfrac{>}{ }\dfrac{3}{2}\left(a,b,c>0\right)\)
Cho các số dương \(a,b,c\) thoả mãn \(a+b+c=3\). Chứng minh rằng: \(\dfrac{a^2+bc}{b+ca}+\dfrac{b^2+ca}{c+ab}+\dfrac{c^2+ab}{a+bc}\ge3\)
cho các số thực dương a,b,c thỏa mãn ab+bc+ca=3.
chứng minh: M=\(\sqrt{\dfrac{bc}{a^2+3}}+\sqrt{\dfrac{ac}{b^2+3}}\sqrt{\dfrac{ab}{c^2+3}}\le\dfrac{3}{2}\)
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) \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
Cho a, b, c là 3 số dương thỏa mãn ab + bc + ca = 3abc. Chứng minh:
\(\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\le\dfrac{3}{2}\)
\(ab+bc+ca=3abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Đặt \(\dfrac{1}{a}=x;\dfrac{1}{b}=y;\dfrac{1}{c}=z\)\(\Rightarrow x+y+z=3\)
\(VT=\sum\dfrac{xyz}{yz+x^2}\le\sum\dfrac{xyz}{2x\sqrt{yz}}=\dfrac{1}{2}\sum\sqrt{yz}\le\dfrac{1}{2}\sum x=\dfrac{3}{2}\)
cho a, b, c \(\ne\) 0. CMR:
\(\dfrac{a^3-b^3}{ab^2}+\dfrac{b^3-c^3}{bc^2}+\dfrac{c^3-a^3}{ca^2}\ge0\)
\(\dfrac{a^3-b^3}{ab^2}+\dfrac{b^3-c^3}{bc^2}+\dfrac{c^3-a^3}{ca^2}\ge0\)
\(\Leftrightarrow\dfrac{a^2}{b^2}-\dfrac{b}{a}+\dfrac{b^2}{c^2}-\dfrac{c}{b}+\dfrac{c^2}{a^2}-\dfrac{a}{c}\ge0\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge\dfrac{2a}{c}\\\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2b}{a}\\\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge\dfrac{2c}{b}\end{matrix}\right.\)
Cộng 3 cái vế theo vế rồi rút gọn cho 2 ta được ĐPCM
Cho a + b + c = 0 và a,b,c \(\ne\) 0.
Chứng minh rằng: \(\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}=-\dfrac{3}{2}\)