Cho a, b, c > 0 và \(a+b+c=1\). Chứng minh: \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\ge\dfrac{1}{3}\)
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>0 chứng minh
\(P=\dfrac{a}{\sqrt{ab+b^2}}+\dfrac{b}{\sqrt{bc+c^2}}+\dfrac{c}{\sqrt{ca+a^2}}\ge\dfrac{3\sqrt{2}}{2}\)
\(\dfrac{P}{\sqrt{2}}=\dfrac{a}{\sqrt{2b\left(a+b\right)}}+\dfrac{b}{\sqrt{2c\left(b+c\right)}}+\dfrac{c}{\sqrt{2a\left(a+c\right)}}\)
\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2a}{2b+a+b}+\dfrac{2b}{2c+b+c}+\dfrac{2c}{2a+a+c}\)
\(\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)=2\left(\dfrac{a^2}{a^2+3ab}+\dfrac{b^2}{b^2+3bc}+\dfrac{c^2}{c^2+3ca}\right)\)
\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)
\(\Rightarrow P\ge\dfrac{3\sqrt{2}}{2}\) (đpcm)
\(\dfrac{a}{\sqrt{ab+b^2}}=\dfrac{\sqrt{2}.a}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{\sqrt{2}.a}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)
làm tương tự với \(\dfrac{b}{\sqrt{bc+c^2}};\dfrac{c}{\sqrt{ca+a^2}}\)
\(=>P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\)
\(=2\sqrt{2}\left(\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)
\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+\dfrac{4}{3}\left(ab+bc+ca\right)+\dfrac{8}{3}\left(ab+bc+ca\right)}\right]\)
\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{\dfrac{4}{3}\left(a+b+c\right)^2}\right]=\dfrac{2\sqrt{2}.3}{4}=\dfrac{3\sqrt{2}}{2}\)
dấu"=" xảy ra<=>a=b=c
Cho 3 số dương a, b, c thỏa mãn: ab+bc+ca=3. Chứng minh: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\)
\(ab+bc+ac=3\)
Ta có:
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) ( đúng với mọi \(ab\ge1\))
Giả sử:\(ab\ge1\)
\(\Rightarrow\dfrac{2}{ab+1}+\dfrac{1}{c^2+1}\ge\dfrac{2c^2+2+ab+1}{\left(ab+1\right)\left(c^2+1\right)}=\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\)
Giả sử: \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\)(đúng)
\(\Leftrightarrow2\left(2c^2+ab+3\right)\ge3\left(ab+1\right)\left(c^2+1\right)\)
\(\Leftrightarrow4c^2+2ab+6\ge3\left(abc^2+ab+c^2+1\right)\)
\(\Leftrightarrow4c^2+2ab+6\ge3abc^2+3ab+3c^2+3\)
\(\Leftrightarrow c^2-ab-3abc^2+3\ge0\)
\(\Leftrightarrow c^2-ab-3abc^2+ab+ac+bc\ge0\) ( vì \(ab+ac+bc=3\) )
\(\Leftrightarrow c^2+ac+bc-3abc^2\ge0\)
\(\Leftrightarrow c+a+b-3abc\ge0\)
\(\Leftrightarrow c+a+b\ge3abc\)
Ta có:
\(3\left(c+a+b\right)=\left(ab+ac+bc\right)\left(c+a+b\right)\) ( vì \(ab+ac+bc=3\) )
Áp dụng BĐT AM-GM, ta có:
\(\left(ab+ac+bc\right)\left(c+a+b\right)\ge9abc\)
\(\Rightarrow a+b+c\ge3abc\)
\(\Rightarrow\) \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\) ( luôn đúng )
\(\Rightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\) ( đfcm )
Dấu "=" xảy ra khi \(a=b=c=1\)
Hình như sai đề rồi bạn ạ, dấu ≥ phải là ≤
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 a+b+c=3.Chứng minh rằng
\(\dfrac{a}{ab+1}+\dfrac{b}{bc+1}+\dfrac{c}{ca+1}\ge\dfrac{3}{2}\)
Chứng minh : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\left(x^2-y^2-xy-xz+2yz\right)^2+\left(y^2-z^2-yz-xy+2xz\right)^2+\left(z^2-x^2-xz-yz+2xy\right)^2\right)\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge a-\dfrac{a^2b}{2\sqrt{ab}}=a-\dfrac{\sqrt{a^3b}}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b}{bc+1}\ge b-\dfrac{\sqrt{b^3c}}{2};\dfrac{c}{ca+1}\ge c-\dfrac{\sqrt{c^3a}}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge3-\dfrac{1}{2}\left(\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
Ta có BĐT \(abc\le\left(\dfrac{a+b+c}{3}\right)^3=1\)
Và \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=3\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{ab+1}+\dfrac{b}{bc+1}+\dfrac{c}{ca+1}\)
\(\ge\dfrac{1}{b+bc}+\dfrac{1}{b+ac}+\dfrac{1}{c+ab}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a+b+c+ab+bc+ca}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
+) Cho a,b,c>0 tm: abc=1
\(CMR:a^3+b^3+c^3+\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}\ge\dfrac{9}{2}\)
Đặt vế trái BĐT cần chứng minh là P, ta có:
\(\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}=\dfrac{1}{c\left(a^2+b^2\right)}+\dfrac{1}{a\left(b^2+c^2\right)}+\dfrac{1}{b\left(c^2+a^2\right)}\)
\(\ge\dfrac{9}{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}\ge\dfrac{9}{2\left(a^3+b^3+c^3\right)}\)
\(\Rightarrow P\ge a^3+b^3+c^3+\dfrac{9}{2\left(a^3+b^3+c^3\right)}\ge3\sqrt[3]{\left(\dfrac{a^3+b^3+c^3}{2}\right)^2.\dfrac{9}{2\left(a^3+b^3+c^3\right)}}\)
\(=3\sqrt[3]{\dfrac{9\left(a^3+b^3+c^3\right)}{8}}\ge3\sqrt[3]{\dfrac{27abc}{8}}=\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a, b, c > 0 thoả mãn: \(a+b+c=1\). Chứng minh: \(\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}\ge\dfrac{3}{2}\)
Hiển nhiên sai em nhé
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\leq \frac{ab}{2ab}+\frac{bc}{2bc}+\frac{ca}{2ca}=\frac{3}{2}\) theo BĐT AM-GM
cho \(a,b,c>0\).CMR
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\ge\dfrac{a+b+c}{3}\)
Áp dụng BĐT AG-GM:
\(\dfrac{a^3}{a^2+ab+b^2}\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}\)
Cmtt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}\\\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\end{matrix}\right.\)
Cộng vế theo vế của bất đẳng thức:
\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)
Tiếp tục áp dụng BĐT AG-GM:
\(\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)
Cmtt\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\\\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\end{matrix}\right.\)
Cộng vế theo vế
\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\\ \ge\dfrac{2}{3}\left(a-\dfrac{b}{2}+b-\dfrac{c}{2}+c-\dfrac{a}{2}\right)=\dfrac{2}{3}\left(a+b+c-\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{3}\)
\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^2.ab.b^2}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)
Tương tự và cộng lại ta sẽ có đpcm
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$