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ÁP dụng BĐT AM-Gm ta có:
\(Σ\frac{a^2}{\left(ab+2\right)\left(2ab+1\right)}\ge\frac{4}{9}\cdotΣ\frac{a^2}{\left(ab+1\right)^2}\)
ĐẶt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\) thì cần cm
\(Σ\frac{a^2}{\left(ab+1\right)^2}=Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{3}{4}\)
\(Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\left(\frac{xz}{y\left(x+z\right)}\right)^2\)
Theo C-S \(Σ\frac{xz}{y\left(x+z\right)}=\frac{\left(xz\right)^2}{xyz\left(x+z\right)}\ge\frac{\left(Σxy\right)^2}{2xy\left(Σx\right)}\ge\frac{3}{2}\)
\(\frac{1}{3}\cdot\left(Σ\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}\)
Đúng hay ta có ĐPCM xyar ra khi a=b=c=1
\(P=\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(b+1\right)\left(c+1\right)}+\frac{c^3}{\left(c+1\right)\left(a+1\right)}-1\)
p \(\ge\)\(\frac{4}{a^2+b^2+2\left(a+b\right)}\) +\(\sqrt{\left(1+ab\right)^2}\) (bunhia và cosi)
=\(\frac{4}{a^2+b^2+2ab}+1+ab=\frac{4}{\left(a+b\right)^2}+a+b+1\)
do \(a+b=ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow a+b\ge4\)
dạt a+b = t thì t>=4
cần tìm min \(\frac{4}{t^2}+t+1=\frac{4}{t^2}+\frac{t}{16}+\frac{t}{16}+\frac{7t}{8}+1\)
\(\ge3.\sqrt[3]{\frac{4}{t^2}.\frac{t}{16}.\frac{t}{16}}+\frac{7.4}{8}+1=\frac{21}{4}\)
dau = xay ra khi a=b=2
Kurosaki Akatsu giải thế thì đề bài cho \(b^2+c^2\le a^2\) để làm gì?
Áp dụng bất đẳng thức AM-GM ta có :
\(P=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(P=\frac{b^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{a^2}{c^2}\ge4.\sqrt[4]{\frac{b^2}{a^2}.\frac{c^2}{a^2}.\frac{a^2}{b^2}.\frac{a^2}{c^2}}=4.1=4\)
=> \(Min_P=4\)
Ta co:
\(P\ge21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{2017.9}{2}\)
\(=21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{18153}{2}\)
\(\Leftrightarrow\frac{P}{\left(a+b+c\right)^2}\ge21\left[\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{a+b+c}\right)^2+\left(\frac{c}{a+b+c}\right)^2\right]+12+\frac{\frac{18153}{2}}{\left(a+b+c\right)^2}\)
Dat \(\left(\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right)\rightarrow\left(x;y;z\right)\)
\(\Rightarrow x+y+z=1\)
\(\Rightarrow\left(a+b+c\right)^2=\frac{a^2}{x^2}\)
BDT tro thanh:
\(\frac{P}{\left(a+b+c\right)^2}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\)
\(\Leftrightarrow\frac{P}{\frac{a^2}{x^2}}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\ge21.\frac{\left(x+y+z\right)^2}{3}+12+\frac{18153}{8}\)
\(\Leftrightarrow\frac{x^2P}{a^2}\ge7+12+\frac{18153}{8}\)
Ta lai co:\(x=\frac{a}{a+b+c}\ge\frac{a}{2}\Rightarrow a^2\le4x^2\)
Suy ra:\(\frac{x^2P}{a^2}\ge\frac{x^2P}{4x^2}=\frac{P}{4}\)
\(\Rightarrow\frac{P}{4}\ge\frac{18503}{8}\)
\(\Leftrightarrow P\ge\frac{18503}{2}\)
Dau '=' xay ra khi \(a=b=c=\frac{2}{3}\)
Vay \(P_{min}=\frac{18503}{2}\)khi \(a=b=c=\frac{2}{3}\)
\(\Rightarrow5a^2+\frac{5}{2}\left(b+c\right)^2\le9a\left(b+c\right)+18bc\le9a\left(b+c\right)+\frac{9}{2}\left(b+c\right)^2\)
\(\Rightarrow5a^2\le9a\left(b+c\right)+2\left(b+c\right)^2\)
\(\Rightarrow5a^2\le18.\frac{a}{2}\left(b+c\right)+2\left(b+c\right)^2\le9\left(\frac{a^2}{4}+\left(b+c\right)^2\right)+2\left(b+c\right)^2\)
\(\Rightarrow\frac{11a^2}{4}\le11\left(b+c\right)^2\Rightarrow b+c\ge\frac{a}{2}\Rightarrow\frac{b+c}{a}\ge\frac{1}{2}\)
\(P\ge1010\left(\frac{b+c}{a}\right)^2+\frac{a}{b+c}\)
Đặt \(\frac{b+c}{a}=x\ge\frac{1}{2}\Rightarrow P\ge1010x^2+\frac{1}{x}=1006x^2+4x^2+\frac{1}{2x}+\frac{1}{2x}\)
\(\Rightarrow P\ge1006.\left(\frac{1}{2}\right)^2+3\sqrt[3]{\frac{4x^2}{4x^2}}=...\)