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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
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
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))
Từ (1) \(\Rightarrow bcx+acy+abz=0\)
Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)
Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)
\(=4\)
\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)
Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)
Ta có
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
Ta có
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)
\(\Rightarrow\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
\(\Rightarrow\frac{2xy.abc^2+2yz.a^2bc+2xz.ab^2c}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
Ta có \(cxy+ayz+bxz=0\)
\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
\(\Rightarrow\frac{2abc.0}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
\(\Rightarrow1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)
bài này bạn bình phương vế thứ 2 lên rồi phân k vế 1 là ra đấy
Đặt : x/a = m ; y/b = n ; z/c = p
=> m+n+p = 1 ; 1/m+1/n+1/p=0
1/m+1/n+1/p=0
<=> mn+np+pm/mnp=0
<=> mn+np+pm=0
<=> 2mn+2np+2pm=0
Xét : 1 = (m+n+p)^2 = m^2+n^2+p^2+2mn+2np+2pm = m^2+n^2+p^2
=> x^2/a^2+y^2/b^2+z^2/c^2 = 1
=> ĐPCM
Tk mk nha
Em(mình) thử nhé, ko chắc đâu
3/ Ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]+\left[ca\left(c+a\right)+ca\right]-abc\)
\(=\left(a+b+c\right)ab+\left(a+b+c\right)bc+\left(a+b+c\right)ca-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)= -abc
Suy ra \(P=\frac{-abc}{abc}=-1\)
Vậy..
