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Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
\(\Rightarrow\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}\cdot\frac{b+c}{4bc}}=\frac{1}{a}\)
\(\Rightarrow\frac{ca}{b^2\left(c+a\right)}+\frac{c+a}{4ca}\ge2\sqrt{\frac{ca}{b^2\left(c+a\right)}\cdot\frac{c+a}{4ca}}=\frac{1}{b}\)
\(\Rightarrow\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}\cdot\frac{a+b}{4ab}}=\frac{1}{c}\)
Cộng theo vế các bất đẳng thức trên ta được:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}+\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Mà\(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)nên:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
hay\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
Bất đẳng thức xảy ra khi \(a=b=c\)
Áp dụng hệ quả BĐT Cauchy cho 2 số thực dương ta có
(ab)^2 +(bc)^2 >=2 ab.bc
(bc)^2+(ca)^2 >= 2bc.ca
(ca)^2+(ab)^2 >= 2ca.ab
=> 2(a^2b^2+b^2c^2+c^2a^2)>=2abc(a+b+c)
<=> a^2b^2+b^2c^2+c^2a^2 >= abc(a+b+c)
Dấu = xảy ra <=> ab=bc=ca <=>a=b=c
Áp dụng bất đẳng thức cosi cho lần lượt 3 số không âm là a,b,c ta có :
\(a^2b^2+b^2c^2\ge2b^2ac\)
\(b^2c^2+c^2a^2\ge2c^2ab\)
\(a^2b^2+c^2a^2\ge2a^2bc\)
Cộng lần lượt 3 vế của các bđt trên ta có :
\(2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\)
ĐPCM
Dấu "=" khi a=b=c
Từ \(a^2b^2+b^2c^2+c^2a^2\ge a^2b^2c^2\)\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\)
bài này tui làm rồi ở đây
a/ Với mọi số thực ta luôn có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Lại có do a;b;c là ba cạnh của 1 tam giác nên theo BĐT tam giác ta có:
\(a+b>c\Rightarrow ac+bc>c^2\)
\(a+c>b\Rightarrow ab+bc>b^2\)
\(b+c>a\Rightarrow ab+ac>a^2\)
Cộng vế với vế: \(2\left(ab+bc+ca\right)>a^2+b^2+c^2\)
b/
Do a;b;c là ba cạnh của tam giác nên các nhân tử vế phải đều dương
Ta có:
\(\left(a+b-c\right)\left(b+c-a\right)\le\frac{1}{4}\left(a+b-c+b+c-a\right)^2=b^2\)
Tương tự: \(\left(a+b-c\right)\left(a+c-b\right)\le a^2\)
\(\left(b+c-a\right)\left(a+c-b\right)\le c^2\)
Nhân vế với vế:
\(a^2b^2c^2\ge\left(a+b-c\right)^2\left(b+c-a\right)^2\left(a+c-b\right)^2\)
\(\Leftrightarrow abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)
Bài 1:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)[a(b+c)+b(c+a)+c(a+b)]\geq (a+b+c)^2\)
\(\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\)$(*)$
Áp dụng BĐT AM-GM dễ thấy: $a^2+b^2+c^2\geq ab+bc+ac$
$\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq \frac{(a+b+c)^2}{3}(**)$
Từ $(*); (**)\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{(a+b+c)^2}{2.\frac{(a+b+c)^2}{3}}=\frac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Áp dụng BĐT AM-GM:
\(\frac{a^3}{b(2c+a)}+\frac{b}{3}+\frac{2c+a}{9}\geq 3\sqrt[3]{\frac{a^3}{b(2c+a)}.\frac{b}{3}.\frac{2c+a}{9}}=a\)
\(\frac{b^3}{c(2a+b)}+\frac{c}{3}+\frac{2a+b}{9}\geq b\)
\(\frac{c^3}{a(2b+c)}+\frac{a}{3}+\frac{2b+c}{9}\ge c\)
Cộng theo vế và thu gọn ta có:
\(\frac{a^3}{b(2c+a)}+\frac{b^3}{c(2a+b)}+\frac{c^3}{a(2b+c)}\geq \frac{a+b+c}{3}=\frac{3}{3}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
