Đặt vế trái là P

\(\frac{a^3}{b^2}+b+b\ge3\sqrt[3]{\frac{a^3b^2}{b^2}}=3a\)

Tương tự: \(\frac{b^3}{c^2}+2c\ge3b\) ; \(\frac{c^3}{d^2}+2d\ge3c\); \(\frac{d^3}{a^2}+2a\ge3d\)

Cộng vế với vế:

\(P+2\left(a+b+c+d\right)\ge3\left(a+b+c+d\right)\)

\(\Leftrightarrow P\ge a+b+c+d\)

Dấu "=" xảy ra khi \(a=b=c=d\)

Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)

Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)

Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)

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

Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*

\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{​​}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)

\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)

Đẳng thức xảy ra khi a = b = c

P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:

1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)

\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)

bài 2 xem có ghi nhầm ko

3a biến đổi tí là xong

b tuong tự bài 1 

Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

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

ta sẽ giết ngươi kí tên dép đờ kiu lờ

Xét BĐT phụ  \(\frac{a^3}{a^2+b^2}\ge\frac{2a-b}{2}\)\(\Leftrightarrow b\left(a-b\right)^2\ge0\)

Tương tự ta có:

\(\frac{b^3}{b^2+c^2}\ge\frac{2b-c}{2};\frac{c^3}{c^2+d^2}\ge\frac{2c-d}{2};\frac{d^3}{d^2+a^2}\ge\frac{2d-a}{2}\)

Cộng lại theo vế ta có:

\(VT\ge\frac{2a-b}{2}+\frac{2b-c}{2}+\frac{2c-d}{2}+\frac{2d-a}{2}\)

\(=\frac{2a-b+2b-c+2c-d+2d-a}{2}=\frac{a+b+c+d}{2}\)

Vậy BĐT đc chứng minh

Áp dụng BĐT cosi ta có

\(\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\); \(\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge\frac{3}{b^2c}\); \(\frac{1}{c^3}+\frac{1}{c^3}+\frac{1}{d^3}\ge\frac{3}{c^2d}\)

\(\frac{1}{d^3}+\frac{1}{d^3}+\frac{1}{a^3}\ge\frac{3}{d^2a}\)

Cộng các BĐt trên ta có 

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\ge\frac{1}{a^2b}+\frac{1}{b^2c}+\frac{1}{c^2d}+\frac{1}{d^2a}\)(1)

Áp dụng BĐT buniacoxki ta có

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

Kết hợp với (1)  ta được ĐPCM

Dấu bằng xảy ra khi a=b=c

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, 

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