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\(\Sigma\frac{1}{6+a}\ge\frac{1}{2}\)
\(\Rightarrow\frac{1}{6+a}\ge\left(\frac{1}{6}-\frac{1}{6+b}\right)+\left(\frac{1}{6}-\frac{1}{6+c}\right)+\left(\frac{1}{6}-\frac{1}{6+d}\right)\)
\(=\frac{b}{6\left(6+b\right)}+\frac{c}{6\left(6+c\right)}+\frac{d}{6\left(6+d\right)}\ge3\sqrt[3]{\frac{bcd}{6\left(6+b\right).6\left(6+c\right).6\left(6+d\right)}}\)
\(=\frac{1}{2}\sqrt[3]{\frac{bcd}{\left(6+b\right)\left(6+c\right)\left(6+d\right)}}\)
Tương tự ta có :
\(\frac{1}{6+b}\ge\frac{1}{2}\sqrt[3]{\frac{acd}{\left(6+a\right)\left(6+c\right)\left(6+d\right)}}\)
\(\frac{1}{6+c}\ge\frac{1}{2}\sqrt[3]{\frac{abd}{\left(6+a\right)\left(6+b\right)\left(6+d\right)}}\)
\(\frac{1}{6+d}\ge\frac{1}{2}\sqrt[3]{\frac{abc}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Nhận các vế với nhau ta được :
\(\frac{1}{\left(6+a\right)\left(6+b\right)\left(6+c\right)\left(6+d\right)}\ge\frac{1}{16}.\sqrt[3]{\left(\frac{abcd}{\left(6+a\right)\left(6+b\right)\left(6+c\right)\left(6+d\right)}\right)^3}\)
\(\Rightarrow\frac{abcd}{16}\le1\)
\(\Rightarrow abcd\le16\)
Dấu " = " xảy ra khi \(a=b=c=d=2\)
Chúc bạn học tốt !!
\(\frac{1}{6+a}\ge\frac{1}{6}-\frac{1}{6+b}+\frac{1}{6}-\frac{1}{6+c}+\frac{1}{6}-\frac{1}{6+c}\)
\(\frac{1}{6+a}\ge\frac{b}{6\left(6+b\right)}+\frac{c}{6\left(6+c\right)}+\frac{d}{6\left(6+d\right)}\ge\frac{1}{2}\sqrt[3]{\frac{bcd}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Tương tự: \(\frac{1}{6+b}\ge\frac{1}{2}\sqrt[3]{\frac{acd}{\left(6+a\right)\left(6+c\right)\left(6+d\right)}}\) ; \(\frac{1}{6+c}\ge\frac{1}{2}\sqrt[3]{\frac{abd}{\left(6+a\right)\left(6+b\right)\left(6+d\right)}}\)
\(\frac{1}{6+d}\ge\frac{1}{2}\sqrt[3]{\frac{abc}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Nhân vế với vế và rút gọn:
\(1\ge\frac{abcd}{16}\Rightarrow abcd\le16\)
Dấu "=" xảy ra khi \(a=b=c=d=2\)
Có: \(VT=\frac{abc}{a^2\left(b+c\right)}+\frac{abc}{b^2\left(c+a\right)}+\frac{abc}{c^2\left(a+b\right)}\)
\(=\frac{bc}{ab+ac}+\frac{ac}{bc+ba}+\frac{ab}{ac+bc}\)
Áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)được
\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{2\left(ab+bc+ca\right)}\)
Mà\(\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2\ge3\left(ab+bc+ca\right)\)(Chuyển vế đưa thành tổng bình phương)
\(\Rightarrow VT\ge...\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" khi a=b=c=1
Cho mk k nhé!
4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13
1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)
Áp dụng bđt Cosi ta có: \(\frac{a^2}{a+b}+\frac{a+b}{4}\ge2;\frac{b^2}{b+c}+\frac{b+c}{4}\ge2;\frac{c^2}{c+d}+\frac{c+d}{4}\ge2\)\(;\frac{d^2}{d+a}+\frac{d+a}{4}\ge2\)
Cộng theo vế và a+b+c+d=1 ta có đpcm
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{a^2}{a+b}=\frac{a+b}{4};\frac{b^2}{b+c}=\frac{b+c}{4};\frac{c^2}{c+d}=\frac{c+d}{4};\frac{d^2}{d+a}=\frac{d+a}{4}\\\\a=b=c=1\end{cases}}\)
\(\Leftrightarrow a=b=c=d=\frac{1}{4}\)
\(\Sigma\frac{1}{6+a}\ge\frac{1}{2}\)
\(\Rightarrow\frac{1}{6+a}\ge\left(\frac{1}{6}-\frac{1}{6+b}\right)+\left(\frac{1}{6}-\frac{1}{6+c}\right)+\left(\frac{1}{6}-\frac{1}{6+d}\right)\)
\(=\frac{b}{6\left(6+b\right)}+\frac{c}{6\left(6+c\right)}+\frac{d}{6\left(6+d\right)}\ge3\sqrt[3]{\frac{bcd}{6\left(6+b\right).6\left(6+c\right).6\left(6+d\right)}}\)
\(=\frac{1}{2}\sqrt[3]{\frac{bcd}{\left(6+b\right)\left(6+c\right)\left(6+d\right)}}\)
tương tự \(\frac{1}{6+b}\ge\frac{1}{2}\sqrt[3]{\frac{acd}{\left(6+a\right)\left(6+c\right)\left(6+d\right)}}\)
\(\frac{1}{6+c}\ge\frac{1}{2}\sqrt[3]{\frac{abd}{\left(6+a\right)\left(6+b\right)\left(6+d\right)}}\)
\(\frac{1}{6+d}\ge\frac{1}{2}\sqrt[3]{\frac{abc}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Nhân các vế lại với nhau đc
\(\frac{1}{\left(6+a\right)\left(6+b\right)\left(6+c\right)\left(6+d\right)}\ge\frac{1}{16}.\sqrt[3]{\left(\frac{abcd}{\left(6+a\right)\left(6+b\right)\left(6+c\right)\left(6+d\right)}\right)^3}\)
\(\Rightarrow\frac{abcd}{16}\le1\)
\(\Rightarrow abcd\le16\)
Dấu "=" tại a = b = c = d = 2