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\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)
\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)
Cộng vế:
\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)
\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\frac{a^3}{(b+2)(c+3)}+\frac{b+2}{36}+\frac{c+3}{48}\geq 3\sqrt[3]{\frac{a^3}{36.48}}=\frac{a}{4}\)
Tương tự:\(\frac{b^3}{(c+2)(a+3)}+\frac{c+2}{36}+\frac{a+3}{48}\geq \frac{b}{4}\)
\(\frac{c^3}{(a+2)(b+3)}+\frac{a+2}{36}+\frac{b+3}{48}\geq \frac{c}{4}\)
Cộng theo vế các BĐT trên và rút gọn ta có:
\(\frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\)
Mà cũng theo AM-GM:
\(a+b+c\geq 3\sqrt[3]{abc}=3\)
\(\Rightarrow \frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\geq \frac{29}{144}.3-\frac{17}{48}=\frac{1}{4}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b+1}{8}+\dfrac{c+1}{8}\)
\(\ge3\sqrt[3]{\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\cdot\dfrac{b+1}{8}\cdot\dfrac{c+1}{8}}=\dfrac{3a}{4}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4};\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\Leftrightarrow VT\ge\dfrac{3}{4}=VP\)
Khi \(a=b=c=1\)
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b}{3}+\dfrac{c+2}{9}\ge3\sqrt[3]{\dfrac{a^3b\left(b+2\right)}{27b\left(c+2\right)}}=a\)
Tương tự: \(\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c}{3}+\dfrac{a+2}{9}\ge b\)
\(\dfrac{c^3}{a\left(b+2\right)}+\dfrac{a}{3}+\dfrac{b+2}{9}\ge c\)
Cộng vế:
\(VT+\dfrac{4\left(a+b+c\right)}{9}+\dfrac{2}{3}\ge a+b+c\)
\(\Rightarrow VT\ge\dfrac{5\left(a+b+c\right)}{9}-\dfrac{2}{3}\ge\dfrac{15}{9}-\dfrac{2}{3}=1\)
Áp dụng BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(\text{luôn đúng}\right)\)
Ta có \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Cộng VTV \(\Leftrightarrow3\ge\dfrac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\sqrt[3]{abc}+1\)
\(\Leftrightarrow VT^2=\sum\left[\dfrac{1}{a\left(b+1\right)}\right]^2\ge3\cdot\sum\dfrac{1}{ab\left(a+1\right)\left(b+1\right)}\\ \Leftrightarrow VT^2\ge3\cdot\dfrac{a^2+b^2+c^2+a+b+c}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\cdot\dfrac{a+b+c+ab+bc+ca}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\\ \Leftrightarrow VT^2\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(1+\sqrt[3]{abc}\right)^3}\\ \Leftrightarrow VT^2\ge\dfrac{9}{\sqrt[3]{\left(abc\right)^2}\left(1+\sqrt[3]{abc}\right)^2}=VP^2\\ \LeftrightarrowĐpcm\)
Dấu \("="\Leftrightarrow a=b=c=1\)
đặt ab=x, bc=y, ac=z
suy ra \(x^3+y^3+z^3=3xyz\)
pt thanh nhân tử \(\left(x+y+z\right)\left(x^2+y^2+z^2-xz-xy-yz\right)=0\)
do x,y,z>0suy ra x+y+z>0
nên suy ra \(x^2+y^2+z^2-xz-yz-xy=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xz-2xy-2yz=0\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
suy ra x=y=z
thế vào pt ta có dpcm
3: \(\left\{{}\begin{matrix}a+b>=2\sqrt{ab}\\b+c>=2\sqrt{bc}\\a+c>=2\sqrt{ac}\end{matrix}\right.\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)>=8abc\)
1: =>(a+b)(a^2-ab+b^2)-ab(a+b)>=0
=>(a+b)(a^2-2ab+b^2)>=0
=>(a+b)(a-b)^2>=0(luôn đúng)
kh có ý 2 à cậu?