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Áp dụng BĐT phụ:
\(3\left(a^2+a^2+b^2\right)\ge\left(2a+b\right)^2\)
P=\(\sum\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P=\sum\dfrac{a}{\sqrt{3\left(a^2+a^2+b^2\right)}+3}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\sum\dfrac{a}{\sqrt{\left(2a+b\right)^2}+a+b+c}=\sum\dfrac{a}{3a+2b+c}\)
Xét M=\(\sum\dfrac{a}{3a+2b+c}\)
\(3-3M=\sum\dfrac{2b+c}{3a+2b+c}\)
\(\Rightarrow\)\(3-3M=\sum\dfrac{\left(2b+c\right)^2}{\left(3a+2b+c\right)\left(2b+c\right)}\ge\)\(\dfrac{\left(3a+3b+3c\right)^2}{\sum\left(3a+2b+c\right)\left(2b+c\right)}\)
Mà
\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)
\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrow\)\(M\le\dfrac{1}{2}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)
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
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^3}{2b+3c}+\frac{b^3}{2c+3a}+\frac{c^3}{2a+3b}=\frac{a^4}{2ab+3ac}+\frac{b^4}{2bc+3ba}+\frac{c^4}{2ac+3bc}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{2ab+3ac+2bc+3ba+2ac+3bc}=\frac{(a^2+b^2+c^2)^2}{5(ab+bc+ac)}\)
Theo hệ quả của BĐT AM-GM ta có:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)(ab+bc+ac)}{5(ab+bc+ac)}=\frac{a^2+b^2+c^2}{5}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(\dfrac{1}{2a^2+b^2}=\dfrac{1}{a^2+a^2+b^2}\le\dfrac{1}{9}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{2b^2+c^2}\le\dfrac{1}{9}\left(\dfrac{1}{b^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\\\dfrac{1}{2c^2+a^2}\le\dfrac{1}{9}\left(\dfrac{1}{c^2}+\dfrac{1}{c^2}+\dfrac{1}{a^2}\right)\end{matrix}\right.\)
Cộng theo vế:
\(L\le\dfrac{1}{9}\left(\dfrac{3}{a^2}+\dfrac{3}{b^2}+\dfrac{3}{c^2}\right)=\dfrac{1}{9}\left[3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right]=\dfrac{1}{9}\)
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 \(a=b=c\)
Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)
\(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)
\(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)
Cộng 3 cái vào, ta có
A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)
Vậy A min = 24
Neetkun ^^