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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 $a=b=c=1$
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
Giải:
Áp dụng BĐT Cauchy cho nhiều số dương:
\(1+\dfrac{1}{a}=\dfrac{a+1}{a}=\dfrac{a+a+b+c}{a}\ge\dfrac{4\sqrt[4]{a^2.b.c}}{a}\)
\(1+\dfrac{1}{b}=\dfrac{b+1}{b}=\dfrac{a+b+b+c}{b}\ge\dfrac{4\sqrt[4]{a.b^2.c}}{a}\)
\(1+\dfrac{1}{c}=\dfrac{c+1}{c}=\dfrac{a+b+c+c}{b}\ge\dfrac{4\sqrt[4]{a.b.c^2}}{c}\)
Nhân vế theo vế, được:
\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge\dfrac{64\sqrt[4]{a^4.b^4.c^4}}{a.b.c}\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge\dfrac{64.abc}{abc}\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
Vậy ...
\(\Leftrightarrow\left(1+abc\right)\left(\dfrac{1}{a\left(1+b\right)}+\dfrac{1}{b\left(1+c\right)}+\dfrac{1}{c\left(1+a\right)}\right)\ge3\)
Ta có:
\(\left(1+abc\right).\dfrac{1}{a\left(1+b\right)}=\dfrac{1+abc}{a+ab}=\dfrac{1+a+ab+abc-a-ab}{a+ab}=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}-1\)
\(\Rightarrow VT=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}+\dfrac{1+b}{b\left(1+c\right)}+\dfrac{c\left(1+a\right)}{1+c}+\dfrac{1+c}{c\left(1+a\right)}+\dfrac{a\left(1+b\right)}{1+a}-3\)
\(VT\ge6\sqrt[6]{\dfrac{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}-3=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
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
abc = 1 nưa nha