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Bài 1: diendantoanhoc.net
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) BĐT cần chứng minh trở thành
\(\frac{x}{\sqrt{3zx+2yz}}+\frac{x}{\sqrt{3xy+2xz}}+\frac{x}{\sqrt{3yz+2xy}}\ge\frac{3}{\sqrt{5}}\)
\(\Leftrightarrow\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}+\frac{y}{\sqrt{5x}\cdot\sqrt{3y+2z}}+\frac{z}{\sqrt{5y}\cdot\sqrt{3z+2x}}\ge\frac{3}{5}\)
Theo BĐT AM-GM và Cauchy-Schwarz ta có:
\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}\ge2\)\( {\displaystyle \displaystyle \sum }\)\(\frac{x}{3x+2y+5z}\ge\frac{2\left(x+y+z\right)^2}{x\left(3x+2y+5z\right)+y\left(5x+3y+2z\right)+z\left(2x+5y+3z\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+7\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(xy+yz+zx\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(x^2+y^2+z^2\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x^2+y^2+z^2\right)}{5\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]}=\frac{3}{5}\)
Bổ sung bài 1:
BĐT được chứng minh
Đẳng thức xảy ra <=> a=b=c
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow\frac{ab+bc+ca}{abc}=1\Rightarrow ab+bc+ca=abc\)\
Ta có: \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=ab+bc+ca-abc=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Từ đây ta suy ra đpcm.
Ta có: \(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)
\(\Leftrightarrow F=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)
\(\Leftrightarrow F\ge\frac{\left(a+b+c+d\right)^2}{2ac+2bd+\left(a+c\right)\left(b+d\right)}=P\)
\(\Leftrightarrow P=\frac{a^2+b^2+c^2+2ab+2bc+2cd+2ad+2ac+2bd}{ab+ac+bc+bd+cd+ac+ad+bd}\)
\(\Leftrightarrow P=\frac{\left(a^2+c^2\right)+\left(b^2+d^2\right)+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+bc+cd+ad}\)
(Vì \(a^2+c^2\ge2ac\Leftrightarrow\left(a-c\right)^2\ge0\)luôn đúng; \(b^2+d^2\ge2bd\Leftrightarrow\left(b-d\right)^2\ge0\)luôn đúng)
\(\Leftrightarrow P\ge\frac{2ac+2bd+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+cd+ad+ac+bd}\)
\(\Leftrightarrow P\ge\frac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\)
\(\Leftrightarrow F\ge P\ge2\)
\(\LeftrightarrowĐPCM\)
1) \(21x^2+21y^2+z^2\)
\(=18\left(x^2+y^2\right)+z^2+3\left(x^2+y^2\right)\)
\(\ge9\left(x+y\right)^2+z^2+3.2xy\)
\(\ge2.3\left(x+y\right).z+6xy\)
\(=6\left(xy+yz+zx\right)=6.13=78\)
Dấu "=" xảy ra <=> x = y ; 3(x+y) = z; xy + yz + zx= 13 <=> x = y = 1; z= 6
2) \(x+y+z=3xyz\)
<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\)
Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)=> ab + bc + ca = 3
Ta cần chứng minh: \(3a^2+b^2+3c^2\ge6\)
Ta có: \(3a^2+b^2+3c^2=\left(a^2+c^2\right)+2\left(a^2+c^2\right)+b^2\)
\(\ge2ac+\left(a+c\right)^2+b^2\ge2ac+2\left(a+c\right).b=2\left(ac+ab+bc\right)=6\)
Vậy: \(\frac{3}{x^2}+\frac{1}{y^2}+\frac{3}{z^2}\ge6\)
Dấu "=" xảy ra <=> a = c = \(\sqrt{\frac{3}{5}}\); \(b=2\sqrt{\frac{3}{5}}\)
khi đó: \(x=z=\sqrt{\frac{5}{3}};y=\sqrt{\frac{5}{3}}\)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Vì a+b+c=0=>(a+b)=-c. Tương tự:(b+c)=-a;(a+c)=-b.
Ta có A=:\(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\frac{a^2}{\left(a-b\right).\left(-c\right)-c^2}+tươngtự\)
\(=\frac{a^2}{-ca+bc-c^2}\)+ tương tự
\(=\frac{a^2}{c\left(b-c-a\right)}+tươngtự\)
\(=\frac{a^2}{c\left(b-\left(c+a\right)\right)}\)+ tương tự nha
\(=\frac{a^2}{c\left(b-\left(-b\right)\right)}+tươngtự=\frac{a^2}{2bc}+tươngtự\)
Sau đó ta có :\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2bc}\)
=\(\frac{a^3+b^3+c^3}{2abc}=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2abc}\)
\(=\frac{\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)}{2abc}\)=\(\frac{0-0-3ab\left(-c\right)}{2abc}\)(do a+b+c=0)
=\(\frac{3abc}{2abc}=\frac{3}{2}\)Ok r bạn
Xét: \(9M=\Sigma\frac{a^2+b^2+c^2}{4a^2+b^2+c^2}-\frac{3}{2}+\Sigma\frac{2\left(ab+bc+ca\right)}{4a^2+b^2+c^2}-3+\frac{9}{2}\)
\(=\Sigma\left(\frac{a^2+b^2+c^2}{4a^2+b^2+c^2}-\frac{1}{2}\right)+\Sigma\left(\frac{2\left(ab+bc+ca\right)}{4a^2+b^2+c^2}-1\right)+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\frac{b^2+c^2-2a^2}{\left(4a^2+b^2+c^2\right)}+\Sigma\frac{2ab+2bc+2ca-4a^2-b^2-c^2}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\frac{\left(b-a\right)\left(b+a\right)+\left(c-a\right)\left(c+a\right)}{\left(4a^2+b^2+c^2\right)}+\Sigma\frac{2a\left[\left(b-a\right)+\left(c-a\right)\right]}{4a^2+b^2+c^2}-\Sigma\frac{\left(b-c\right)^2}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\left(\frac{\left(a-b\right)\left(a+b\right)}{a^2+4b^2+c^2}-\frac{\left(a-b\right)\left(b+a\right)}{4a^2+b^2+c^2}\right)-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\left(a-b\right)\left(a+b\right)\left(\frac{3a^2-3b^2}{\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}\right)-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\Sigma\frac{3\left(a-b\right)^2\left(a+b\right)^2}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\Sigma\left(a-b\right)^2\left[\frac{3\left(a+b\right)^2}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{1}{a^2+b^2+4c^2}\right]-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\Sigma\left(a-b\right)^2\left[\frac{3\left(a+b\right)^2\left(a^2+b^2+4c^2\right)-2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(a^2+b^2+4c^2\right)}\right]-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}+\frac{9}{2}\)Ai đó làm tiếp giúp em vs:( Em chỉ nghĩ ra được tới đây thôi.
Ta có:
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;a^2+c^2\ge2\sqrt{a^2c^2}=2ac;a^2+a^2\ge2\sqrt{a^2a^2}=2a^2\)
Khi đó:
\(4a^2+b^2+c^2\ge2a\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{4a^2+b^2+c^2}\le\frac{1}{6a}\)
Tương tự:
\(\frac{1}{a^2+4b^2+c^2}\le\frac{1}{6b};\frac{1}{a^2+b^2+4c^2}\le\frac{1}{6c}\cdot\)
\(\Rightarrow M\le\frac{1}{6}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{ab+bc+ca}{abc}\cdot\frac{1}{6}\) \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow3\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)
Theo BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Khi đó \(M\le\frac{3}{1}\cdot\frac{1}{6}=\frac{1}{2}\)
Dấu "=" xảy ra tại \(a=b=c=1\)
P/S:Is that true ??
Ta có: abcd=1 và a+b+c+d=\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\)
Do đó: a+b-\(\left(\frac{1}{a}+\frac{1}{b}\right)+c+d-\left(\frac{1}{c}+\frac{1}{d}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(1-\frac{1}{ab}\right)+\left(c+d\right)\left(1-\frac{1}{cd}\right)=0\)
\(\Leftrightarrow\frac{\left(a+b\right)\left(ab-1\right)}{ab}+\left(c+d\right)\left(1-ab\right)=0\)
\(\Leftrightarrow\left(ab-1\right)\left(\frac{a+b}{ab}-c-d\right)=0\)
\(\Leftrightarrow\left(ab-1\right)\left(a+b-abc-abd\right)=0\)
\(\Leftrightarrow\left(ab-1\right)\left[a\left(1-bc\right)+b\left(1-ad\right)\right]=0\)
\(\Leftrightarrow\left(ab-1\right)\left[a\left(1-bc\right)+b\left(abcd-ad\right)\right]=0\)
\(\Leftrightarrow\left(ab-1\right)\left(1-bc\right)\left(a-abd\right)=0\)
\(\Leftrightarrow a\left(ab-1\right)\left(1-bc\right)\left(1-bd\right)=0\)
<=> ab-1=0 hoặc 1-bc=0 hoặc 1-bd=0
<=> ab=1 hoặc bc=1 hoặc bd=1
\(\Leftrightarrow a\left(ab-1\right)\left(1-bc\right)\left(1-bd\right)=0\)