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Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
Áp dụng BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)
Ta chỉ cần chứng minh:
\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)
Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)
Nên ta chỉ cần chứng minh:
\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
\(VT=\dfrac{a^2}{a+abc}+\dfrac{b^2}{b+abc}+\dfrac{c^2}{c+abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\dfrac{1}{9}\left(a+b+c\right)^3}=\dfrac{1^2}{1+\dfrac{1}{9}.1^3}=\dfrac{9}{10}\)
Đặt \(a=\dfrac{yz}{x^2};b=\dfrac{zx}{y^2};c=\dfrac{xy}{z^2}\)
Áp dụng BĐT BSC:
\(\dfrac{1}{a^2+a+1}+\dfrac{1}{b^2+b+1}+\dfrac{1}{c^2+c+1}\)
\(=\dfrac{x^4}{x^4+x^2yz+y^2z^2}+\dfrac{y^4}{y^4+y^2zx+z^2x^2}+\dfrac{z^4}{z^4+z^2xy+x^2y^2}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)}\)
Ta cần chứng minh:
\(\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)}\ge1\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\ge x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2-xy.yz-yz.zx-zx.xy\ge0\)
\(\Leftrightarrow\left(xy-yz\right)^2+\left(yz-zx\right)^2+\left(zx-xy\right)^2\ge0,\forall x,y,z\)
\(\Rightarrow dpcm\)
Đẳng thức xảy ra khi \(a=b=c=1\)
\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)
\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)
Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)
\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)
\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)
\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)
Đặt \(a\left(1-b\right)=x;b\left(1-c\right)=y;c\left(1-a\right)=x\)
\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca=1-a\left(1-b\right)-b\left(1-c\right)-c\left(1-a\right)=1-x-y-z\)
BĐT cần c/m trở thành:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{3}{1-x-y-z}\)
\(\Leftrightarrow\left(1-x-y-z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-3\ge0\)
\(\Leftrightarrow\dfrac{1-x-y-z}{x}+\dfrac{1-x-y-z}{y}+\dfrac{1-x-y-z}{z}-3\ge0\)
\(\Leftrightarrow\dfrac{1-y-z}{x}+\dfrac{1-z-x}{y}+\dfrac{1-x-y}{z}-6\ge0\) (1)
Lại có: \(1-y-z=1-b\left(1-c\right)-c\left(1-a\right)=1-b-c+bc+ca=\left(1-b\right)\left(1-c\right)+ca\)
Nên (1) tương đương:
\(\dfrac{\left(1-b\right)\left(1-c\right)+ca}{a\left(1-b\right)}+\dfrac{\left(1-a\right)\left(1-c\right)+ab}{b\left(1-c\right)}+\dfrac{\left(1-a\right)\left(1-b\right)+bc}{c\left(1-a\right)}-6\ge0\)
\(\Leftrightarrow\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\)
BĐT trên hiển nhiên đúng theo AM-GM do:
\(\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\sqrt[6]{\dfrac{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}}=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Cám ơn bài giải của thầy Lâm ạ!
Và từ bài bất đăng thức này, đã được chế thành bài toán hình học trong 1 kì thi học sinh giỏi toán cấp tỉnh thầy ạ!
Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)
Đơn giản là kiên nhẫn tính toán và tách biểu thức:
\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)
Sau đó Cô-si cho từng ngoặc là được
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\((a+b+1)(a+b+c^2)\geq (a+b+c)^2\Rightarrow a+b+1\geq \frac{(a+b+c)^2}{a+b+c^2}\)
\(\Rightarrow \frac{1}{a+b+1}\leq \frac{a+b+c^2}{(a+b+c)^2}\)
Tương tự cho các phân thức còn lại, suy ra:
\(1\leq \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{a+c+1}\leq \frac{a+b+c^2}{(a+b+c)^2}+\frac{b+c+a^2}{(a+b+c)^2}+\frac{c+a+b^2}{(a+b+c)^2}\)
\(\Leftrightarrow 1\leq \frac{2(a+b+c)+a^2+b^2+c^2}{(a+b+c)^2}\)
\(\Leftrightarrow (a+b+c)^2\leq 2(a+b+c)+a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac\leq a+b+c\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=1$