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Bất đẳng thức cần chứng minh tương đương với:
\(a^3b^2-a^2b^3+b^3c^2-c^3b^2+c^3a^2-c^2a^3\ge0\)
\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\ge0\)
\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b+b-a\right)\ge0\)
\(\Leftrightarrow a^2b^2\left(a-b\right)+c^2a^2\left(b-a\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b\right)\ge0\)
\(\Leftrightarrow\left(a^2b^2-c^2a^2\right)\left(a-b\right)+\left(b^2c^2-c^2a^2\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow a^2\left(b^2-c^2\right)\left(a-b\right)+c^2\left(b^2-a^2\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow\left[a^2\left(b+c\right)-c^2\left(a+b\right)\right]\left(a-b\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow\left(a^2b+a^2c-c^2a-c^2b\right)\left(a-b\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow\left[a\left(ab-c^2\right)+c\left(a^2-bc\right)\right]\left(a-b\right)\left(b-c\right)\ge0\) luôn đúng do \(a\ge b\ge c\ge0\)
cảm ơn bạn nhá, bạn trả lời giúp mình mấy câu hỏi về BĐT còn lại của mik đc ko? cảm ơn bn nhiều!
Ta có:
A = \(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{3b+2a}=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{3bc+2ac}\)
A \(\ge\frac{\left(a+b+c\right)^2}{2ab+3ac+2bc+3ab+3bc+2ac}\)(bđt svacxo \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\))
A \(\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ac\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{5\left(a+b+c\right)^2}{3}}\) (bđt \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)(*)
CM bđt * <=> \(3xy+3yz+3xz\le x^2+y^2+z^2+2xz+2xy+2yz\)
<=> \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\ge0\) (luôn đúng)
<=> A \(\ge\frac{3}{5}\) --> ĐPCM
Áp dụng bđt Cauchy-Schwarz:
\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\ge\frac{\left(1+1+1\right)^2}{2a+b+c+a+2b+c+a+b+2c}=\frac{9}{4a+4b+4c}\)Dấu "=" xảy ra khi a=b=c
\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)
\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)
\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)
Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:
\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)
\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)
\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)
Cộng (1),(2) và (3) có:
\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)
\(\Rightarrow2VP\ge2VT\)
\(\RightarrowĐPCM\)
a) Áp dụng bất đẳng thức AM-GM :
\(\left(a^2+b^2\right)\left(a^2+1\right)\ge2\sqrt{a^2b^2}.2\sqrt{a^2}\ge2ab.2a=4a^2b\)
b) Áp dụng bất đẳng thức :\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x;y>0\)
\(\frac{1}{a+3b}+\frac{1}{b+2c+a}\ge\frac{4}{a+3b+b+2c+a}=\frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
Tương tự \(\hept{\begin{cases}\frac{1}{b+3c}+\frac{1}{c+2a+b}\ge\frac{2}{b+2c+a}\\\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{2}{b+2a+c}\end{cases}}\)
Cộng vế với vế ta được : \(VT+VP\ge2VP\Rightarrow VT\ge VP\)(đpcm)
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
\(a^3+a^3+b^3\ge3\sqrt[3]{a^6b^3}=3a^2b\)
\(b^3+b^3+c^3\ge3\sqrt[3]{b^6c^3}=3b^2c\)
\(c^3+c^3+a^3\ge3\sqrt[3]{c^6a^3}=3c^2a\)
Cộng vế theo vế có ngay điều phải chứng minh
\(a^5+a^5+a^5+a^5+b^5\ge5\sqrt[5]{a^{20}b^5}=5a^4b\)
\(b^5+b^5+b^5+b^5+c^5\ge5\sqrt[5]{b^{20}c^5}=5b^4c\)
\(c^5+c^5+c^5+c^5+a^5\ge5\sqrt[5]{c^{20}a^5}=5c^4a\)
Cộng lại ta được:\(5\left(a^5+b^5+c^5\right)\ge5\left(a^4b+b^4c+c^4a\right)\)
=> đpcm