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\(P=\dfrac{4a^2}{4b+2c}+\dfrac{4b^2}{4a+2c}+\dfrac{c^2}{4a+4b}\ge\dfrac{\left(2a+2b+c\right)^2}{8a+8b+4c}\)
\(=\dfrac{\left(2a+2b+c\right)^2}{4\left(2a+2b+c\right)}=\dfrac{1}{4}\left(2a+2b+c\right)\)
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
Dấu BĐT bị ngược, sửa đề: \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).
Đặt \(b^2=x\left(x>0\right)\Rightarrow a+x=2ax\).
Khi đó ta cần chứng minh:
\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)
Áp dụng BĐT AM-GM:
\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\)
\(\le\dfrac{1}{2a^2x+2ax^2}+\dfrac{1}{2ax^2+2a^2x}\)
\(=\dfrac{2}{2ax\left(a+x\right)}\)
\(=\dfrac{1}{ax\left(a+x\right)}\)
\(=\dfrac{1}{2a^2x^2}\)
Ta thấy: \(a+x\ge2\sqrt{ax}\)
\(\Leftrightarrow2ax\ge2\sqrt{ax}\)
\(\Leftrightarrow ax-\sqrt{ax}\ge0\)
\(\Leftrightarrow\sqrt{ax}\left(\sqrt{ax}-1\right)\ge0\)
\(\Leftrightarrow\sqrt{ax}\ge1\)
\(\Rightarrow ax\ge1\)
Khi đó: \(\dfrac{1}{2a^2x^2}\le\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)
Hay \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).
Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).
\(\Leftrightarrow\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}-2+\dfrac{c\left(2b-c\right)}{b\left(c+a\right)}-2+\dfrac{a\left(2c-a\right)}{c\left(a+b\right)}-2\le\dfrac{3}{2}-6\)
\(\Leftrightarrow\dfrac{b^2+2ac}{a\left(b+c\right)}+\dfrac{c^2+2ab}{b\left(c+a\right)}+\dfrac{a^2+2bc}{c\left(a+b\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{b^2}{ab+ac}+\dfrac{c^2}{bc+ab}+\dfrac{a^2}{ac+bc}+\dfrac{2c^2}{bc+c^2}+\dfrac{2a^2}{ac+a^2}+\dfrac{2b^2}{ab+b^2}\ge\dfrac{9}{2}\)
Ta có:
\(VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}+\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)
\(\Leftrightarrow VT\ge\left(a+b+c\right)^2\left(\dfrac{1}{2\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}\right)\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+2\left(a^2+b^2+c^2+ab+bc+ca\right)}\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=\dfrac{9}{2}\)