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\(\Sigma\left(\frac{a^3}{a^2+b^2}\right)=\Sigma\left(\frac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}\right)=\Sigma\left(a-\frac{ab^2}{a^2+b^2}\right)\ge\Sigma\left(a-\frac{ab^2}{2ab}\right)=\Sigma\left(a-\frac{b}{2}\right)\)
\(=a+b+c-\left(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)=\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Theo bất đẳng thức Cauchy-Schwarzt ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}.\)
Mặt khác, \(a^2+b^2+c^2\ge ab+bc+ca\), do đó ta suy ra \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2.\)
P=\(\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=a^2+b^2+c^2\)
bđt \(\Leftrightarrow\)\(\Sigma_{cyc}\frac{a^2+ab+ca}{\left(b+c\right)^2}\ge\frac{9}{4}\)
Có: \(\frac{a^2+ab+ca}{\left(b+c\right)^2}=\frac{a^2+ab+bc+ca}{\left(b+c\right)^2}-\frac{bc}{\left(b+c\right)^2}\ge\frac{\left(a+b\right)\left(c+a\right)}{\left(b+c\right)^2}-\frac{1}{4}\)
=> \(\Sigma_{cyc}\frac{a^2+ab+ca}{\left(b+c\right)^2}\ge3\sqrt[3]{\frac{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}}-\frac{3}{4}=\frac{9}{4}\)
Áp dụng BĐT Svác ta có:
\(\frac{a^2}{2b+c}+\frac{b^2}{2c+a}+\frac{c^2}{2a+b}\ge\frac{\left(a+b+c\right)^2}{3\left(a+b+c\right)}=\frac{a+b+c}{3}\)
2. Bạn kiểm tra lại đề: VP = 1/2
Ta có:
\(\sqrt{a\left(3a+b\right)}=\frac{1}{4}.2.\sqrt{4a\left(3a+b\right)}\le\frac{1}{4}\left(4a+3a+b\right)=\frac{1}{4}\left(7a+b\right)\)
\(\sqrt{b\left(3b+a\right)}=\frac{1}{4}.2.\sqrt{4b\left(3b+a\right)}\le\frac{1}{4}\left(4b+3b+a\right)=\frac{1}{4}\left(7b+a\right)\)
=> \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{\frac{1}{4}\left(7a+b\right)+\frac{1}{4}\left(7b+a\right)}=\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
Vậy: \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{1}{2}\) với a, b dương
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel :
\(\text{∑}\frac{a}{b+c}=\text{∑}\frac{a^2}{ab+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)(1)
Bạn chứng minh bđt \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(2)
Từ (1) và (2) \(\Rightarrow\text{∑}\frac{a}{b+c}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\left(đpcm\right)\)
Dấu "=" xảy ra <=> a = b = c