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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)
1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(\sqrt[3]{a+3b}=\sqrt[3]{1.1.(a+3b)}\leq \frac{1+1+a+3b}{3}\)
\(\Rightarrow \frac{1}{\sqrt[3]{a+3b}}\geq \frac{3}{a+3b+2}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
$\Rightarrow P\geq 3\left(\frac{1}{a+3b+2}+\frac{1}{b+3c+2}+\frac{1}{c+3a+2}\right)$
Áp dụng BĐT Cauchy- Schwarz:
\(\frac{1}{a+3b+2}+\frac{1}{b+3c+2}+\frac{1}{c+3a+2}\geq \frac{9}{4(a+b+c)+6}=\frac{9}{4.\frac{3}{4}+6}=1\)
Do đó: $P\geq 3.1=3$
Vậy $P_{\min}=3$ khi $a=b=c=\frac{1}{4}$
Ta có : \(\frac{bc}{\sqrt{3a+bc}}=\frac{bc}{\sqrt{\left(a+b+c\right)a+bc}}=\frac{bc}{\sqrt{a^2+ab+ac+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cauchy , ta có : \(\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
Tương tự : \(\frac{ac}{\sqrt{3b+ac}}=\frac{ac}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{ac}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\); \(\frac{ab}{\sqrt{3c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{ab}{2}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(\Rightarrow P=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{ac}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{ab}{\sqrt{\left(a+c\right)\left(c+b\right)}}\)
\(\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
Suy ra : Max P \(=\frac{3}{2}\Leftrightarrow a=b=c=1\)
đây nhé Câu hỏi của Steffy Han - Toán lớp 8 | Học trực tuyến
1) Áp dụng BĐT bunhia, ta có
\(P^2\le3\left(6a+6b+6c\right)=18\Rightarrow P\le3\sqrt{2}\)
Dấu = xảy ra <=> a=b=c=1/3