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Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)
\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)
\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ab+ac}{abc}=2\)
\(\frac{bc+ab+ac}{a+b+c}=2\Leftrightarrow bc+ab+ac=2\left(a+b+c\right)\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}\)( * )
Để \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)thì \(2\left(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}\right)=2\Leftrightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=1\)
\(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{a^2bc+bac^2+ab^2c}{\left(abc\right)^2}=\frac{abc\left(a+b+c\right)}{\left(abc\right)^2}=\frac{a+b+c}{abc}\)
mà a + b + c = abc \(\Rightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{abc}{abc}=1\Leftrightarrow\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\)
thay \(\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\) vào ( * ) ta được \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\left(đpcm\right)\)
\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{bc.ac+ab.ac+ab.bc}{ab.bc.ac}\)
\(=\frac{abc.\left(a+b+c\right)}{a^2b^2c^2}=\frac{a+b+c}{abc}=1\left(\text{vì }a+b+c=abc\right)\)
\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\text{ từ}\left(1\right)\)
Vậy ...
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
"Chấm" nhẹ hóng cao nhân ạ :)
P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)
Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath
Học tốt=)
tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2
1 .
Từ gt : \(2ab+6bc+2ac=7abc\)và \(a,b,c>0\)
Chia cả hai vế cho abc > 0
\(\Rightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)
Đặt \(x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}\Rightarrow\hept{\begin{cases}x,y,z>0\\2z+6x+2y=7\end{cases}}\)
Khi đó : \(C=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}\)
\(=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)
\(\Rightarrow C=\frac{4}{2x+y}+2x+y+\frac{9}{4x+z}+4x+z+\frac{4}{y+z}+y+z\)\(-\left(2x+y+4x+z+y+z\right)\)
\(=\left(\frac{2}{\sqrt{x+2y}}-\sqrt{x+2y}\right)^2+\left(\frac{3}{\sqrt{4x+z}}-\sqrt{4x+z}\right)^2\)\(+\left(\frac{2}{\sqrt{y+z}}-\sqrt{y+z}\right)^2+17\ge17\)
Khi \(x=\frac{1}{2},y=z=1\)thì \(C=17\)
Vậy GTNN của C là 17 khi a =2; b =1; c = 1
2 .
Áp dụng bất đẳng thức Cauchy ta có :\(1+b^2\ge2b\)nên
\(\frac{a+1}{1+b^2}=\left(a+1\right)-\frac{b^2\left(a+1\right)}{b^2+1}\)
\(\ge\left(a+1\right)-\frac{b^2\left(a+1\right)}{2b}=a+1-\frac{ab+b}{2}\)
\(\Leftrightarrow\frac{a+1}{1+b^2}\ge a+1-\frac{ab+b}{2}\left(1\right)\)
Tương tự ta có:
\(\frac{b+1}{1+c^2}\ge b+1-\frac{bc+c}{2}\left(2\right)\)
\(\frac{c+1}{1+a^2}\ge c+1-\frac{ca+a}{2}\left(3\right)\)
Cộng vế theo vế (1), (2) và (3) ta được:
\(\frac{a+1}{1+b^2}+\frac{b+1}{1+c^2}+\frac{c+1}{1+a^2}\ge3+\frac{a+b+c-ab-bc-ca}{2}\left(^∗\right)\)
Mặt khác : \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2=9\)
\(\Rightarrow\frac{a+b+c-ab-bc-ca}{2}\ge0\)
Nên \(\left(^∗\right)\) \(\Leftrightarrow\frac{a+1}{1+b^2}+\frac{b+1}{1+c^2}+\frac{c+1}{1+a^2}\ge3\left(đpcm\right)\)
Dấu " = " xảy ra khi và chỉ khi \(a=b=c=1\)
Chúc bạn học tốt !!!
1 bai thoi cung dc