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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
2,
a, Nếu 2a + 4 \(\ge\) 2b + 4
thì 2a \(\ge\) 2b hay a \(\ge\) b
b, Nếu 3a - 5 \(\le\) 3b - 5
thì 3a \(\le\) 3b hay a \(\le\) b
3,
a, Nếu a \(\le\) b thì a - b \(\le\) 0 hay 2019(a - b) \(\le\) 0 hay 2019a \(\le\) 2019b hay 2019a + 2020 \(\le\) 2019b + 2020
b, Nếu a \(\le\) b thì -a \(\ge\) -b hay -42a \(\ge\) -42b hay -42a - 24 \(\ge\) -42b - 24
3,
a, Nếu a > b thì 3a > 3b hay 3a + 2 > 3b + 2
b, Nếu a > b thì -a < -b hay -4a < -4b hay -4a - 5 < -4b - 5
Chúc bn học tốt!!
3. Câu hỏi của Hoàng Đức Thịnh - Toán lớp 8 - Học toán với OnlineMath
\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)
\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)
\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)
`a/(2b+3c) +b/(2c+3a) + c/(2a+3b) >=3/5`
Thiếu đk `a,b,c>0`
`a/(2b+3c) +b/(2c+3a) + c/(2a+3b)`
`=a^2/(2ab+3ac)+b^2/(2bc+3ab)+c^2/(2ac+3bc)`
Áp dụng BĐT cosi-schwart:
`a^2/(2ab+3ac)+b^2/(2bc+3ab)+c^2/(2ac+3bc)>=(a+b+c)^2/(5(ab+bc+ca))=(a^2+b^2+c^2+2ab+2bc+2ca)//(5(ab+bc+ca))`
Áp dụng cosi:`a^2+b^2+c^2>=ab+bc+ca`
`=>a^2/(2ab+3ac)+b^2/(2bc+3ab)+c^2/(2ac+3bc)>=(3(ab+bc+ca))/(5(ab+bc+ca))=3/5`
Dấu "=" xảy ra khi `a=b=c`
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
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 \(a=b=c=d\)
Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).
\(a^3+a^3+b^3\ge3\sqrt[3]{a^6b^3}=3a^2b\)
\(b^3+b^3+c^3\ge3\sqrt[3]{b^6c^3}=3b^2c\)
\(c^3+c^3+a^3\ge3\sqrt[3]{c^6a^3}=3c^2a\)
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