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\(BDT\Leftrightarrow\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\ge\dfrac{1}{2a+b+c}+\dfrac{1}{2b+c+a}+\dfrac{1}{2c+a+b}\)
Áp dụng BĐT \(\dfrac{1}{nht}+\dfrac{1}{is}+\dfrac{1}{the}+\dfrac{1}{best}\ge\dfrac{16}{nht+is+the+best}\):
\(\dfrac{1}{2a+b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VP\le\dfrac{4}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\)
\("="\Leftrightarrow a=b=c\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2a+b+c}=\dfrac{a}{a+b+c+a}\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\\\dfrac{b}{a+2b+c}=\dfrac{b}{a+b+b+c}\le\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{c}{a+b+2c}=\dfrac{c}{a+c+b+c}\le\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{a}{4\left(a+b\right)}+\dfrac{a}{4\left(a+c\right)}+\dfrac{b}{4\left(a+b\right)}+\dfrac{b}{4\left(b+c\right)}+\dfrac{c}{4\left(a+c\right)}+\dfrac{c}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\left[\dfrac{a}{4\left(a+b\right)}+\dfrac{b}{4\left(a+b\right)}\right]+\left[\dfrac{b}{4\left(b+c\right)}+\dfrac{c}{4\left(b+c\right)}\right]+\left[\dfrac{c}{4\left(a+c\right)}+\dfrac{a}{4\left(a+c\right)}\right]\)
\(\Rightarrow VT\le\dfrac{a+b}{4\left(a+b\right)}+\dfrac{b+c}{4\left(b+c\right)}+\dfrac{c+a}{4\left(c+a\right)}\)
\(\Rightarrow VT\le\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}\le\dfrac{3}{4}\) ( đpcm )
Dấu "=" xảy ra khi \(a=b=c\)
Lời giải:
Ta có: \(a^2b+b^2c+c^2a\geq \frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)
\(\Leftrightarrow (a^2b+b^2c+c^2a)(1+2a^2b^2c^2)\geq 9a^2b^2c^2\)
\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)(*)\)
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Áp dụng BĐT AM-GM ta có:
\(a^2b+a^4b^3c^2+a^3b^2c^4\geq 3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)
\(b^2c+a^2b^4c^3+a^4b^3c^2\geq 3a^2b^3c^2\)
\(c^2a+a^3b^2c^4+a^2b^4c^3\geq 3a^2b^2c^3\)
Cộng theo vế:
\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)\)
Vậy $(*)$ đúng
Do đó ta có đpcm
Dấu bằng xảy ra khi $a=b=c=1$
Bạn tham khảo lời giải tại đây:
Câu hỏi của Phác Chí Mẫn - Toán lớp 9 | Học trực tuyến
Ta có: \(a^2+2b+3=a^2+2b+1+2\ge2\left(a+b+1\right)\)
Tương tự ta được: \(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b+1}+\dfrac{b}{b+c+1}+\dfrac{c}{c+a+1}\right)\)
Ta sẽ chứng minh \(\dfrac{a}{a+b+1}+\dfrac{b}{b+c+1}+\dfrac{c}{c+a+1}\le1\)
\(\Leftrightarrow\dfrac{-b-1}{a+b+1}+\dfrac{-c-1}{b+c+1}+\dfrac{-a-1}{c+a+1}\le-2\)
\(\Leftrightarrow\dfrac{b+1}{a+b+1}+\dfrac{c+1}{b+c+1}+\dfrac{a+1}{c+a+1}\ge2\)
\(\Leftrightarrow\dfrac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\dfrac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}+\dfrac{\left(a+1\right)^2}{\left(a+1\right)\left(c+a+1\right)}\ge2\left(1\right)\)
Cần chứng minh BĐT (1) đúng
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(a+b+c+3\right)^2}{a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3}\)
Mà \(a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3\)
\(=\dfrac{1}{2}\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+9\right]\)
\(=\dfrac{1}{2}\left(a+b+c+3\right)^2\)\(\Rightarrow VT\left(1\right)\ge2=VP\left(1\right)\)
Đẳng thức xảy ra khi \(a=b=c=1\)
Bđt cauchy-schwarz dạng engel dạng tổng quát là j vây c