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C/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\) (*)
Thật vậy , (*) \(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)
\(\Leftrightarrow ab+bc+ac+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ac\right)+4\left(a+b+c\right)+8\)
\(\Leftrightarrow ab+bc+ac+abc=4\) (Đ)
=> (*) đúng ( đpcm )
Đặt \(P=\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{a+d}+\dfrac{d}{a+b}\)
\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{ac+cd}+\dfrac{d^2}{ad+bd}\)
\(P\ge\dfrac{\left(a+b+c+d\right)^2}{ab+2ac+bc+2bd+cd+ad}=\dfrac{\left(a+c\right)^2+\left(b+d\right)^2+2\left(a+c\right)\left(b+d\right)}{2ac+2bd+ab+bc+cd+ad}\)
\(P\ge\dfrac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d\)
\(VT=\sqrt{\left(a+\dfrac{5b}{2}\right)^2+\dfrac{15b^2}{4}}+\sqrt{\left(b+\dfrac{5c}{2}\right)^2+\dfrac{15c^2}{4}}+\sqrt{\left(c+\dfrac{5a}{2}\right)^2+\dfrac{15a^2}{4}}\)
\(\Rightarrow VT\ge\sqrt{\left(a+\dfrac{5b}{2}+b+\dfrac{5c}{2}+c+\dfrac{5a}{2}\right)^2+\dfrac{15}{4}\left(a+b+c\right)^2}\)
\(\Rightarrow VT\ge\sqrt{\dfrac{49}{4}\left(a+b+c\right)^2+\dfrac{15}{4}\left(a+b+c\right)^2}=4\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:
\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\dfrac{1}{2}.2xy\left(x^2+y^2\right)=xy\left(x^2+y^2\right)\)
Áp dụng:
\(P\le\dfrac{a}{a+bc\left(b^2+c^2\right)}+\dfrac{b}{b+ca\left(c^2+a^2\right)}+\dfrac{c}{c+ab\left(a^2+b^2\right)}\)
\(P\le\dfrac{a^2}{a^2+abc\left(b^2+c^2\right)}+\dfrac{b^2}{b^2+abc\left(c^2+a^2\right)}+\dfrac{c^2}{c^2+abc\left(a^2+b^2\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Đặt vế trái của BĐT cần chứng minh là P
Ta có:
\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+2b+3c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\right)\)
Tương tự:
\(\dfrac{1}{b+2c+3a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\right)\) ; \(\dfrac{1}{c+2a+3b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\right)\)
Cộng vế:
\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
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 \(a=b=c\)