\(3a^2+8b^2+14ab\le3a^2+8b^2+12ab+a^2+b^2=\left(2a+3b\right)^2\)

\(\Rightarrow\sqrt{3a^2+8b^2+14ab}\le2a+3b\)

\(\Rightarrow P=\sum\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\sum\frac{a^2}{2a+3b}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

Chứng minh BĐT phụ: \(\frac{m^2}{x}+\frac{n^2}{y}\ge\frac{\left(m+n\right)^2}{x+y}\) với \(x;y>0\)         (*)

Ta có: \(3a^2+8b^2+14ab\)

\(=\left(3a^2+12ab\right)+\left(2ab+8b^2\right)\)

\(=3a\left(a+4b\right)+2b\left(a+4b\right)\)

\(=\left(3a+2b\right)\left(a+4b\right)\)

\(\Rightarrow\sqrt{3a^2+8b^2+14ab}=\sqrt{\left(3a+2b\right)\left(a+4b\right)}\le\frac{3a+2b+a+4b}{2}=2a+3b\)

\(\Rightarrow\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\frac{a^2}{2a+3b}\)

Tương tự, ta có:  \(\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}\ge\frac{b^2}{2b+3c}\)

                           \(\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\ge\frac{c^2}{2c+3a}\)

Áp dụng (*), ta có:

\(VT\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{2a+3b+2b+3c+2c+3a}=\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}\)

                                                                                         \(=\frac{1}{5}\left(a+b+c\right)\)

Vậy \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\ge\frac{1}{5}\left(a+b+c\right)\)

có thể là bé hơn hoặc bằng,các bạn thử cho mình với nhé

áp dụng Bất Đẳng Thức CBS \(\sqrt{3a^2+8b^2+14ab}=\sqrt{\left(a+4b\right)\left(3a+2b\right)}\le\frac{1}{2}\left(4a+6b\right)\)

(BĐT CBS) do đó ta \(\Rightarrow\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\frac{a^2}{2a+3b}\)

tương tư với mẫu còn lại 

\(\Rightarrow\Sigma\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\Sigma\frac{a^2}{2a+3b}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\left(Q.E.D\right)\)

đẳng thức xảy ra khi a=b=c

Ta có: 

\(\sqrt{3a^2+8b^2+14ab}=\sqrt{\left(3a+2b\right)\left(a+4b\right)}\le2a+3b\)

Khi đó \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\frac{a^2}{2a+3b}\), tương tự ta có:

\(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)

\(\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\)\(\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)

\(\frac{a^2}{\sqrt{3a^2+8b^2+12ab+2ab}}\ge\frac{a^2}{\sqrt{3a^2+9b^2+12ab+a^2+b^2}}=\frac{a^2}{\sqrt{\left(2a+3b\right)^2}}=\frac{a^2}{2a+3b}\)

\(\Rightarrow VT\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{1}{5}\left(a+b+c\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

Ta có: \(\sqrt{3a^2+14ab+8b^2}=\sqrt{\left(2a+3b\right)^2-\left(a-b\right)^2}\)

\(\le\sqrt{\left(2a+3b\right)^2}=2a+3b\)

Tương tự, ta có: \(\sqrt{3b^2+14bc+8c^2}\le2b+3c\); \(\sqrt{3c^2+14ca+8a^2}\le2c+3a\)

\(\Rightarrow\frac{a^2}{\sqrt{3a^2+14ab+8b^2}}+\frac{b^2}{\sqrt{3b^2+14bc+8c^2}}+\frac{c^2}{\sqrt{3c^2+14ca+8a^2}}\)

\(\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)(Theo BĐT Bunyakovski dạng phân thức)

Đẳng thức xảy ra khi a = b = c

\(3a^2+8b^2+2ab+12ab\le3a^2+8b^2+a^2+b^2+12ab=\left(2a+3b\right)^2\)

\(\Rightarrow A\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}=404\)

\(A_{min}=404\) khi \(a=b=c=\frac{2020}{3}\)

Điều kiện là a, b, c>0

Ta phân tích mẫu:

\(\sqrt{3a^2+8b^2+14ab}=\sqrt{\left(3a+2b\right)\left(a+4b\right)}\le\frac{\left(4a+6b\right)}{2}=2a+3b\)

Áp dụng BĐT Cauchy Schwarz, ta có: \(VT\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{\left(a+b+c\right)}{5}\) 

Dấu "=" xảy ra khi a=b=c