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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:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
Lời giải:
Ta thấy:
\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)
\(=\frac{a(a+b+c)}{a+b}+\frac{b(a+b+c)}{b+c}+\frac{c(a+b+c)}{c+a}\)
\(=(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)
Có:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$
$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$
Do đó:
$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)
Áp dụng BĐT AM-GM:
\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)
\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{\sqrt{3}}$
Ta thấy: \(\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}=\Sigma_{cyc}\frac{a^2+bc}{\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}}\)
Ta lại có: \(\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}\le\frac{\left(a^2b+b^2c\right)+\left(bc^2+ca^2\right)+\left(c^2a+ab^2\right)}{3}=\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
\(\Leftrightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{\Sigma_{cyc}\left(a^2+bc\right)}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{a^2+b^2+c^2+ab+bc+ca}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}\)
Nhận thấy: \(A=\left(a+b+c\right)\left(a^2+b^2+c^2+ab+bc+ca\right)=a^3+b^3+c^3+3abc+2\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
Theo Schur: \(a^3+b^3+c^3+3abc\ge\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
\(\Leftrightarrow A\ge3\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{3\Sigma_{cyc}\left(ab\left(a+b\right)\right)}{\frac{1}{3}\left(a+b+c\right)\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{9}{a+b+c}\)
Hình như đề bài có vấn đề : thừa đk ab + bc + ac = abc
ta có : \(\frac{\sqrt{b^2+2a^2}}{ab}\ge\frac{\sqrt{4a^2b^2}}{ab}=\frac{2ab}{ab}=2\)
Tương tự \(\frac{\sqrt{c^2+2b^2}}{bc}\ge2\) ; \(\frac{\sqrt{a^2+2c^2}}{ac}\ge2\)
\(\Rightarrow\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ac}\ge2+2+2=6>\sqrt{3}\)
\(\left(\frac{a}{\sqrt{bc}};\frac{b}{\sqrt{ca}};\frac{c}{\sqrt{ab}}\right)=\left(x;y;z\right)\)\(\Rightarrow\)\(xyz=1\)
\(VT=\frac{1}{x+3}+\frac{1}{y+3}+\frac{1}{z+3}\)
\(VT-\frac{3}{4}=\frac{-3\left(x+y+z\right)-5\left(xy+yz+zx\right)-3xyz+27}{4\left(x+3\right)\left(y+3\right)\left(z+3\right)}\le\frac{-3.3-5.3-3+27}{4\left(x+3\right)\left(y+3\right)\left(z+3\right)}=0\)
sai đề