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Áp dụng BĐT AM - GM ta có:
$ \frac{a^3}{(1 + b)(1 + c)} + \frac{1 + b}{8} + \frac{1 + c}{8} \geq \frac{3}{4}a$
$\frac{b^3}{(1 + c)(1 + a)} + \frac{1 + c}{8} + \frac{1 + a}{8} \geq \frac{3}{4}b$
$\frac{c^3}{(1 + a)(1 + b)} + \frac{1 + a}{8} + \frac{1 + b}{8} \geq \frac{3}{4}c $
Cộng vế theo vế ta được:
$ P + \frac{2(a + b + c) + 6}{8} \geq \frac{3}{4}(a + b + c) $
$<=> P \geq \frac{1}{2}(a + b + c) - \frac{3}{4}$
$=> P \geq \frac{3}{4} (dpcm)$
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a\left(b+c\right)}{4}\ge2\sqrt{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a\left(b+c\right)}{4}=2\sqrt{\dfrac{1}{4a^2}=\dfrac{1}{a}=\dfrac{abc}{a}=bc}}\)
Tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b\left(c+a\right)}{4}\ge\dfrac{1}{b}=ac\)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c\left(a+b\right)}{4}\ge\dfrac{1}{c}=ab\)
Cộng theo vế:
\(\Rightarrow VT+\dfrac{ab+bc+ac}{2}\ge ab+bc+ac\)
\(\Rightarrow VT\ge\dfrac{ab+bc+ac}{2}\)
Tiếp tục áp dụng AM-GM: \(ab+bc+ac\ge3^3\sqrt{a^2b^2c^2}=3\)
\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Dấu bằng xảy ra khi a=b=c=1
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{c+1}{4}+\dfrac{b}{2}\ge3\sqrt[3]{\dfrac{a^3}{b\left(c+1\right)}\cdot\dfrac{c+1}{4}\cdot\dfrac{b}{2}}\)
\(=3\sqrt[3]{\dfrac{a^3}{4\cdot2}\cdot\dfrac{c+1}{c+1}\cdot\dfrac{b}{b}}=3\sqrt[3]{\dfrac{a^3}{8}}=\dfrac{3a}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{c\left(a+1\right)}\ge\dfrac{3b}{2};\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3c}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{a+b+c+3}{4}+\dfrac{a+b+c}{2}\ge\dfrac{3a+3b+3c}{2}\)
\(\Leftrightarrow VT+\dfrac{3\left(a+b+c\right)}{4}+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{2}\)
\(\Leftrightarrow VT+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\). Mà theo AM-GM ta có:
\(a+b+c\ge3\sqrt[3]{abc}=3\)\(\Rightarrow VT+\dfrac{3}{4}\ge\dfrac{9}{4}\Rightarrow VT\ge\dfrac{3}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=1\)
Áp dụng BĐT holder cho n bộ 3 số:
\(\left(\sum\dfrac{b^nc^n}{b+c}\right)\left[\sum\left(b+c\right)\right]\left(1+1+1\right)..\left(1+1+1\right)\ge\left(ab+bc+ca\right)^n\)
\(\Leftrightarrow VT\ge\dfrac{\left(ab+bc+ca\right)^n}{3^{n-2}.2.\left(a+b+c\right)}\ge\dfrac{3^{n-2}.3abc\left(a+b+c\right)}{3^{n-2}.2.\left(a+b+c\right)}=\dfrac{3}{2}\)
#Hint:(\(\left\{{}\begin{matrix}ab+bc+ca\ge3\\\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\end{matrix}\right.\))
BĐT holder thường dùng:
\(\left(a_1^m+a_2^m+...+a_k^m\right)\left(b_1^m+b_2^m+...+b_k^m\right)...\left(c_1^m+...+c_k^m\right)\ge\left(a_1b_1...c_1+a_2.b_2...c_2+...+a_k.b_k...c_k\right)^m\)
trong đó VT có m thừa số từ a đến c
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^4}{(a+2)(b+2)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\geq 4\sqrt[4]{\frac{a^4}{27.27.9}}=\frac{4a}{9}\)
\(\frac{b^4}{(b+2)(c+2)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\geq \frac{4b}{9}\)
\(\frac{c^4}{(c+2)(a+2)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\geq \frac{4c}{9}\)
Cộng theo vế và rút gọn:
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}+\frac{2(a+b+c)}{27}+\frac{7}{9}\geq\frac{4(a+b+c)}{9}\)
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}\geq \frac{10(a+b+c)}{27}-\frac{7}{9}=\frac{30}{27}-\frac{7}{9}=\frac{1}{3}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
@Akai Haruma