Ta có:

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

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

= \(4-\left(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}+\frac{d^2}{d^2+1}\right)\)

Áp dụng Cô - si:

\(a^2+1\ge2\sqrt{a^2.1}=2a\) <=> \(\frac{a^2}{a^2+1}\le\frac{a}{2}\)

Tương tự => \(\left\{{}\begin{matrix}\frac{b^2}{b^2+1}\le\frac{b}{2}\\\frac{c^2}{c^2+1}\le\frac{c}{2}\\\frac{d^2}{d^2+1}\le\frac{d}{2}\end{matrix}\right.\)

<=> \(4-\left(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}+\frac{d^2}{d^2+1}\right)\)

\(\ge4-\frac{a+b+c+d}{2}=2\)

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

3a) ta có \(\frac{a^2}{a+b}=a-\frac{ab}{a+b}>=a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)

vì \(a,b>0,a+b>=2\sqrt{ab}nên\frac{ab}{a+b}< =\frac{ab}{2\sqrt{ab}}\)

tương tự \(\frac{b^2}{b+c}=b-\frac{bc}{b+c}>=b-\frac{bc}{2\sqrt{bc}}=b-\frac{\sqrt{bc}}{2}\)

tương tự \(\frac{c^2}{c+a}=c-\frac{ca}{c+a}>=c-\frac{ca}{2\sqrt{ca}}=c-\frac{\sqrt{ca}}{2}\)

cộng từng vế BĐT ta được \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=a+b+c-\frac{\sqrt{ab}}{2}-\frac{\sqrt{bc}}{2}-\frac{\sqrt{ca}}{2}=\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}\left(1\right)\)

giả sử \(\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}>=\frac{a+b+c}{2}\)

<=> \(2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=a+b+c\)

<=> \(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=0\)

<=> \(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}>=0\)

<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)

(đúng với mọi a,b,c >0) (2)

(1),(2)=> \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=\frac{a+b+c}{2}\left(đpcm\right)\)

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 

Phạm Minh Quang Vũ Minh Tuấn kudo shinichi Lê Thị Thục Hiền Akai Haruma Nguyễn Huy Thắng Nguyễn Thị Diễm Quỳnh Băng Băng 2k6 giúp với ạ mình cần gấp lắm