Ta có: \(P=\frac{25a^2}{\sqrt{2a^2+16ab+7b^2}}+\frac{25b^2}{\sqrt{2b^2+16bc+7c^2}}+\frac{c^2\left(3+a\right)}{a}\)\(=\frac{25a^2}{\sqrt{\left(2a+3b\right)^2-2\left(a-b\right)^2}}+\frac{25b^2}{\sqrt{\left(2b+3c\right)^2-2\left(b-c\right)^2}}+\frac{c^2\left(3+a\right)}{a}\)\(\ge\frac{25a^2}{2a+3b}+\frac{25b^2}{2b+3c}+\frac{c^2\left(3+a\right)}{a}\)

Áp dụng bất đẳng thức AM - GM, ta có: \(\frac{25a^2}{2a+3b}+\left(2a+3b\right)\ge2\sqrt{\frac{25a^2}{2a+3b}.\left(2a+3b\right)}=10a\Rightarrow\frac{25a^2}{2a+3b}\ge8a-3b\)(1)

\(\frac{25b^2}{2b+3c}+\left(2b+3c\right)\ge2\sqrt{\frac{25b^2}{2b+3c}.\left(2b+3c\right)}=10b\Rightarrow\frac{25b^2}{2b+3c}\ge8b-3c\)(2)

\(\frac{c^2\left(3+a\right)}{a}=\frac{3c^2}{a}+c^2=\left(\frac{3c^2}{a}+3a\right)+\left(c^2+1\right)-3a-1\)\(\ge2\sqrt{\frac{3c^2}{a}.3a}+2c-3a-1=8c-3a-1\)(3)

Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(\frac{25a^2}{2a+3b}+\frac{25b^2}{2b+3c}+\frac{c^2\left(3+a\right)}{a}\ge5\left(a+b+c\right)-1=14\)

Vậy \(P\ge14\)

Đẳng thức xảy ra khi a = b = c = 1

Bài 2 :

Ta có :

\(2a^2+16ab+7b^2=\left(2a+3b\right)^2-2\left(a-b\right)^2\le\left(2a+3b\right)^2\)

\(\Rightarrow P\ge\frac{25a^2}{2a+3b}+\frac{25b^2}{2b+3c}+\frac{c^2\left(a+3\right)}{a}\)

Áp dụng BĐT Cô - si ta có :

\(\frac{25a^2}{2a+3b}+2a+3b\ge10a\)

\(\frac{25b^2}{2b+3c}+2b+3c\ge10b\)

\(\frac{c^2\left(a+3\right)}{a}=\left(c^2+1\right)+\left(\frac{3c^2}{a}+3a\right)-3a-1\ge2c+6c-3a-1=8c-3a-1\)

Khi đó :

\(P\ge\left(10-2a-3b\right)+\left(10b-2b-3c\right)+\left(8c-3a-1\right)\)

\(\Rightarrow P\ge5\left(a+b+c\right)-1=14\)

Vậy \(MinP=14\) khi a=b=c=1

Ta có: \(4ab\le2a^2+2b^2\)

=> \(\sqrt{2a^2+7b^2+16ab}\le\sqrt{4a^2+9b^2+12ab}=\sqrt{\left(2a+3b\right)^2}=2a+3b\)

=> \(\frac{a^2}{\sqrt{2a^2+7b^2+16ab}}\ge\frac{a^2}{2a+3b}\)

Chứng minh tương tự 

=> \(T\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\)

Áp dụng bđt bunhia dạng phân thức

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

=> \(MinT=1\)xảy ra khi a=b=c=5/3

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

bài 1

ÁP dụng AM-GM ta có:

\(\frac{a^3}{b\left(2c+a\right)}+\frac{2c+a}{9}+\frac{b}{3}\ge3\sqrt[3]{\frac{a^3.\left(2c+a\right).b}{b\left(2c+a\right).27}}=a.\)

tương tự ta có:\(\frac{b^3}{c\left(2a+b\right)}+\frac{2a+b}{9}+\frac{c}{3}\ge b,\frac{c^3}{a\left(2b+c\right)}+\frac{2b+c}{9}+\frac{a}{3}\ge c\)

công tất cả lại ta có:

\(P+\frac{2a+b}{9}+\frac{2b+c}{9}+\frac{2c+a}{9}+\frac{a+b+c}{3}\ge a+b+c\)

\(P+\frac{2\left(a+b+c\right)}{3}\ge a+b+c\)

Thay \(a+b+c=3\)vào ta được":

\(P+2\ge3\Leftrightarrow P\ge1\)

Vậy Min là \(1\)

dấu \(=\)xảy ra khi \(a=b=c=1\)