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Áp dụng bất đẳng thức Svác xơ ngược ta có
\(\frac{1}{2a+3b+3c}=\frac{1}{a+b+a+c+2\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{2}{b+c}\right)\)
tương tự mấy cái kia rồi cộng vào
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
Thay \(a+b+c=3\) ta được:
\(VT=\frac{1}{a\left(a+b+c\right)+bc}+\frac{1}{b\left(a+b+c\right)+ca}+\frac{1}{c\left(a+b+c\right)+ab}\)
\(=\frac{1}{a^2+ab+ac+bc}+\frac{1}{b^2+ab+bc+ca}+\frac{1}{c^2+ca+bc+ab}\)
\(=\frac{1}{a\left(a+b\right)+c\left(a+b\right)}+\frac{1}{b\left(a+b\right)+c\left(a+b\right)}+\frac{1}{c\left(a+c\right)+b\left(a+c\right)}\)
\(=\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\)
\(=\frac{b+c+a+c+a+b}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{2\left(a+b+c\right)}{\sqrt{\left[\left(a+b\right)\left(a+c\right)\right].\left[\left(a+b\right)\left(b+c\right)\right].\left[\left(a+c\right)\left(b+c\right)\right]}}\)
\(=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}=VP\) (Do \(a+b+c=3\))
=> ĐPCM.
1.
Ta có: \(\frac{2a+3b+3c+1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2ac-1}{2017+c}\)
\(=\frac{b+c+4033}{2015+a}+\frac{c+a+4032}{2016+b}+\frac{a+b+4031}{2017+c}\)
Đặt \(\hept{\begin{cases}2015+a=x\\2016+b=y\\2017+c=z\end{cases}}\)
\(P=\frac{b+c+4033}{2015+a}+\frac{c+a+4032}{2016+b}+\frac{a+b+4031}{2017+c}\)
\(=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}=\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\)
\(\ge2\sqrt{\frac{y}{x}\cdot\frac{x}{y}}+2\sqrt{\frac{z}{x}\cdot\frac{x}{z}}+2\sqrt{\frac{y}{z}\cdot\frac{z}{y}}\left(Cosi\right)\)
Dấu "=" <=> x=y=z => \(\hept{\begin{cases}a=673\\b=672\\c=671\end{cases}}\)
Vậy Min P=6 khi a=673; b=672; c=671
Câu 1 thử cộng 3 vào P xem
Rồi áp dụng BDT Cauchy - Schwars : a^2/x + b^2/y + c^2/z ≥(a + b + c)^2/(x + y + z)
\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)
\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)
\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)
\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)
\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
Vậy VT = VP, đẳng thức được chứng minh
Làm tạm vào đây vậy
từ gt dễ dàng => \(ab+bc+ca\le3\)
\(\Rightarrow\frac{ab}{\sqrt{c^2+3}}\le\frac{ab}{\sqrt{c^2+ab+bc+ca}}=\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng cô si ta có
\(\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}\right)\)
Tương tự như vậy rồi ccộng vào nhá nhok