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\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\ge\frac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự: \(\sqrt{\frac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\) ; \(\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge ca+2b^2\)
Cộng vế với vế:
\(VT\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)
4.
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Dấu "=" xảy ra khi \(a=b=c\)
5.
\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)
Cộng vế với vế:
\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
1.
Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)
\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2.
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)
Cộng vế với vế:
\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3.
Từ câu b, thay \(c=1\) ta được:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
Ta co:
\(VT=\Sigma_{cyc}\frac{a}{ca+1}=\Sigma_{cyc}\frac{a}{ca+abc}=\Sigma_{cyc}\frac{1}{c+bc}\)
Xet
\(\Sigma_{cyc}\frac{1}{c+bc}\le\frac{1}{4}\Sigma_{cyc}\left(\frac{1}{c}+\frac{1}{bc}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{4}\left(ab+bc+ca+a+b+c\right)\)
bdt can chung minh thanh
\(ab+bc+ca+a+b+c\le2\left(a^2+b^2+c^2\right)\)
Ta lai co:
\(a^2+b^2+c^2\ge ab+bc+ca\)
Gio ta can chung minh:
\(a^2+b^2+c^2\ge a+b+c\)
Ta co hai danh gia:
\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)
\(1=\sqrt[3]{abc}\le\frac{a+b+c}{3}\le\frac{\sqrt{3\left(a^2+b^2+c^2\right)}}{3}\Rightarrow a^2+b^2+c^2\ge3\)
Suy ra can chung minh:
\(a^2+b^2+c^2\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2-3\right)\ge0\) (đúng)
Dau '=' xay ra khi \(a=b=c=1\)
mn giup voi minh can gap lam
Vũ Minh TuấnBăng Băng 2k6Nguyễn Việt LâmPhạm Lan HươngNguyễn Huy Tú Nguyễn Thị Thùy TrâmNo choice teentthbảo phạmHo Nhat Minh
\(A\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(A\ge\frac{1}{a^2+b^2+c^2}+\frac{4}{2ab+2ac+2bc}+\frac{7}{ab+bc+ca}\)
\(A\ge\frac{\left(1+2\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}\)
\(A\ge\frac{9}{\left(a+b+c\right)^2}+\frac{21}{\left(a+b+c\right)^2}=30\)
\(A_{min}=30\) khi \(a=b=c=\frac{1}{3}\)