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a)Do bd>0 (do b>0, d>0) nên nếu \(\frac{a}{b}< \frac{c}{d}\) thì ad<bc
b)Ngược lại, nếu ad<bc thì \(\frac{ad}{bd}< \frac{bc}{bd}\Leftrightarrow\frac{a}{b}< \frac{c}{d}\)
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c-a\right)\left(a+b\right)\left(b+c\right)-d\left(c-a\right)\left(c+d\right)\left(d+a\right)=0\)
\(\Leftrightarrow b\left(a+b\right)\left(b+c\right)-d\left(c+d\right)\left(d+a\right)=0\)
\(\Leftrightarrow bad+bd^2+bca+bcd-dab-dac-db^2-cbd=0\)
\(\Leftrightarrow bca-dca+bd^2-db^2=0\)
\(\Leftrightarrow\left(b-d\right)\left(ca-bd\right)=0\)
\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)
\(a,\frac{-2,6}{x}=-\frac{12}{42}\)
\(\Leftrightarrow\left(-2,6\right).42=-12x\)
\(\Leftrightarrow-12x=-\frac{546}{5}\)
\(\Leftrightarrow x=\frac{91}{10}\)
\(b,\frac{x^2}{6}=\frac{24}{25}\)
\(\Leftrightarrow25x^2=24.6\)
\(\Leftrightarrow25x^2=144\)
\(\Leftrightarrow x^2=\frac{144}{25}\)
\(\Leftrightarrow x=\frac{12}{5}\)
\(bdt\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}+\frac{d-a}{a+b}\ge0\)
\(\Leftrightarrow\left(\frac{a-b}{b+c}+1\right)+\left(\frac{b-c}{c+d}+1\right)+\left(\frac{c-d}{d+a}+1\right)+\left(\frac{d-a}{a+b}+1\right)\ge4\)
\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{a+c}{d+a}+\frac{b+d}{a+b}\ge4\)
\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\)(*)
Theo Cauchy-Schwarz:
\(\frac{1}{b+c}+\frac{1}{d+a}\ge\frac{4}{a+b+c+d};\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{a+b+c+d}\)
Khi đó:\(\left(\cdot\right)\ge\left(a+c\right).\frac{4}{a+b+c+d}+\left(b+d\right).\frac{4}{a+b+c+d}=4\)
\(\frac{a^4}{a^3+2b^3}=a-\frac{2ab^3}{a^3+b^3+b^3}\ge a-\frac{2ab^3}{3\sqrt[3]{a^3.b^3.b^3}}=a-\frac{2}{3}b\)
Tương tự ta có
\(\frac{b^4}{b^3+2c^3}\ge b-\frac{2}{3}c\) ; \(\frac{c^4}{c^3+2d^3}\ge c-\frac{2}{3}d\) ; \(\frac{d^4}{d^3+2a^3}\ge d-\frac{2}{3}a\)
Cộng vế với vế:
\(VT\ge a+b+c+d-\frac{2}{3}\left(a+b+c+d\right)=\frac{a+b+c+d}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d\)
Ta có : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
1) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{b}{d}=\frac{a}{c}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
đpcm
2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
đpcm
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
a) \(ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\)
b) \(ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}\)
c) \(ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\)