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Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a+1}\ge1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}+1-\dfrac{1}{d+1}\)
\(=\dfrac{b}{b+1}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\)\(\ge3\sqrt[3]{\dfrac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\)
Tương tự cho 3 BĐT còn lại cũng có:
\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}};\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}};\dfrac{1}{d+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Nhân theo vế 4 BĐT trên ta có:
\(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\sqrt[3]{\left(\dfrac{abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\right)^3}\)
\(\Leftrightarrow1\ge81abcd\Leftrightarrow abcd\le\dfrac{1}{81}\)
\(\dfrac{1}{a^3}+a\ge2\sqrt{\dfrac{a}{a^3}}=\dfrac{2}{a}\) ; \(\dfrac{1}{b^3}+b\ge\dfrac{2}{b}\) ; \(\dfrac{1}{c^3}+c\ge\dfrac{2}{c}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (1)
Lại có \(\dfrac{4a}{a^4+1}\le\dfrac{4a}{2\sqrt{a^4}}=\dfrac{4a}{2a^2}=\dfrac{2}{a}\)
Tương tự \(\dfrac{4b}{b^4+1}\le\dfrac{2}{b}\) ; \(\dfrac{4c}{c^4+1}\le\dfrac{2}{c}\)
\(\Rightarrow4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\le2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (2)
Từ (1),(2)\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)
Dấu "=" xảy ra khi a=b=c=1
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{\left(1+1+1+1\right)^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
Chỉ bằng các kiến thức cho trong SGK (bất đẳng thức Cô si cho hai số không âm; bất đẳng thức Bunhiacopxki cho 2 cặp số) có thể giả bài toán như sau:
Ta có \(\left(a+b+c+d\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)=\)
\(=a\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)+b\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)+d\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
\(=4+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{a}{d}+\dfrac{d}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{d}+\dfrac{d}{b}\right)+\left(\dfrac{c}{d}+\dfrac{d}{c}\right)\)
\(\ge4+2+2+2+2+2+2=16\)
Từ đó \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{16}{a+b+c+d}\). Đẳng thức xảy ra khi và chỉ khi \(\dfrac{a}{b}=\dfrac{b}{a};\dfrac{a}{c}=\dfrac{c}{a};\dfrac{a}{d}=\dfrac{d}{a};\dfrac{b}{c}=\dfrac{c}{b};...\Leftrightarrow a=b=c=d\)
ta có \(\dfrac{1}{\left(a+b\right)c}\le\dfrac{1}{2\sqrt{ab}c}=\dfrac{1}{2\sqrt{c}}\)tương tự ta có
\(\Sigma\dfrac{1}{\left(a+b\right)c}\le\Sigma\dfrac{1}{2\sqrt{c}}=\dfrac{\Sigma\sqrt{ab}}{2}\le\dfrac{\Sigma a}{2}\)(đpcm)
Đặt \(\left\{{}\begin{matrix}x=a-\dfrac{1}{2}\\y=b-\dfrac{1}{2}\\z=c-\dfrac{1}{2}\\t=d-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow x+y+z+t=0\)
\(BDT\Leftrightarrow\dfrac{2\left(2x+1\right)}{4x^2+3}+\dfrac{2\left(2y+1\right)}{4y^2+3}+\dfrac{2\left(2z+1\right)}{4z^2+3}+\dfrac{2\left(2t+1\right)}{4t^2+3}\le\dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)^2}{4x^2+3}+\dfrac{\left(2y-1\right)^2}{4y^2+3}+\dfrac{\left(2z-1\right)^2}{4z^2+3}+\dfrac{\left(2t-1\right)^2}{4t^2+3}\ge\dfrac{4}{3}\left(1\right)\)
Ta có: \(4x^2+3=3x^2+3+\left(y+z+t\right)^2\le3x^2+3+3\left(y^2+z^2+t^2\right)\)
\(=3\left(x^2+y^2+z^2+t^2+1\right)\)
\(\Rightarrow\dfrac{\left(2x-1\right)^2}{4x^2+3}\ge\dfrac{\left(2x-1\right)^2}{3\left(x^2+y^2+z^2+t^2+1\right)}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT_{\left(1\right)}\ge\dfrac{\left(2x-1\right)^2+\left(2y-1\right)^2+\left(2z-1\right)^2+\left(2t-1\right)^2}{3\left(x^2+y^2+z^2+t^2+1\right)}\)
\(=\dfrac{4\left(x^2+y^2+z^2+t^2+1\right)-4\left(x+y+z+t\right)}{3\left(x^2+y^2+z^2+t^2+1\right)}\)
\(=\dfrac{4\left(x^2+y^2+z^2+t^2+1\right)}{3\left(x^2+y^2+z^2+t^2+1\right)}=\dfrac{4}{3}=VP_{\left(1\right)}\)
ta có : \(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{ab}{bc}\right)\left(1+\dfrac{c}{a}\right)\)
\(=1+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{bc}{ac}+\dfrac{a}{b}+\dfrac{ac}{ba}+\dfrac{ab}{bc}+1\)
\(=2+\left(\dfrac{c}{a}+\dfrac{ab}{bc}\right)+\left(\dfrac{b}{c}+\dfrac{ac}{ba}\right)+\left(\dfrac{a}{b}+\dfrac{bc}{ac}\right)\ge2+2+2+2=8\) \(\Rightarrowđpcm\)