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Chứng minh : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\left(x^2-y^2-xy-xz+2yz\right)^2+\left(y^2-z^2-yz-xy+2xz\right)^2+\left(z^2-x^2-xz-yz+2xy\right)^2\right)\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge a-\dfrac{a^2b}{2\sqrt{ab}}=a-\dfrac{\sqrt{a^3b}}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b}{bc+1}\ge b-\dfrac{\sqrt{b^3c}}{2};\dfrac{c}{ca+1}\ge c-\dfrac{\sqrt{c^3a}}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge3-\dfrac{1}{2}\left(\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
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 \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)
\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)
\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)
\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)
Xảy ra khi \(a=b=c\)
\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)
\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)
Giả sử \(a\ge b\ge c\)thì
\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh
\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)
\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)
\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)
\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))
Vậy BĐT được chứng minh.
Dấu = xảy ra khi a=b=c
Bài toán cơ bản:
\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)
Bunhiacopxki:
\(\left(a+b+c\right)\left(\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right)\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\right)^2=1\)
\(\Rightarrow\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\ge\dfrac{1}{a+b+c}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
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!
a.
Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
b.
Ta có:
\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)
Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
Cộng vế với vế:
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
\(\Rightarrow ab+bc+ca=abc\)
Xét \(\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\)
\(\Leftrightarrow\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
\(\Leftrightarrow\dfrac{a^3}{a^2+ab+bc+ca}+\dfrac{b^3}{b^2+ab+bc+ca}+\dfrac{c^3}{c^2+ab+bc+ca}\)
\(\Leftrightarrow\dfrac{a^3}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b^3}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{c^3}{c\left(b+c\right)+a\left(b+c\right)}\)
\(\Leftrightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\\\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3b}{4}\\\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3b}{4}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{a+b+c}{2}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{a+b+c}{2}\)
\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{a+b+c}{4}\)
\(\Leftrightarrow\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\ge\dfrac{a+b+c}{4}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=3\)
p/s: bài này em nhớ em đã giải cho anh ròi mà ta =))
đài thế cách tui ngắn hơn nhiều