Áp dụng BĐT phụ:

\(3\left(a^2+a^2+b^2\right)\ge\left(2a+b\right)^2\)

P=\(\sum\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}\)

\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P=\sum\dfrac{a}{\sqrt{3\left(a^2+a^2+b^2\right)}+3}\)

\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\sum\dfrac{a}{\sqrt{\left(2a+b\right)^2}+a+b+c}=\sum\dfrac{a}{3a+2b+c}\)

Xét M=\(\sum\dfrac{a}{3a+2b+c}\)

\(3-3M=\sum\dfrac{2b+c}{3a+2b+c}\)

\(\Rightarrow\)\(3-3M=\sum\dfrac{\left(2b+c\right)^2}{\left(3a+2b+c\right)\left(2b+c\right)}\ge\)\(\dfrac{\left(3a+3b+3c\right)^2}{\sum\left(3a+2b+c\right)\left(2b+c\right)}\)

Mà

\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)

\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)

\(\Rightarrow\)\(M\le\dfrac{1}{2}\)

\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)

\(\dfrac{4}{3}y=\dfrac{4x+\dfrac{8}{3}}{\sqrt{2x+1}+1}=\dfrac{4x+2+\dfrac{2}{3}}{\sqrt{2x+1}+1}\)

\(\Rightarrow\)\(\dfrac{4}{3}y=\dfrac{\left(\sqrt{2x+1}+1\right)\left(\sqrt{2x+1}-1\right)+\dfrac{2}{3}}{\sqrt{2x+1}+1}=\sqrt{2x+1}-1+\dfrac{2}{\dfrac{3}{\sqrt{2x+1}+1}}\)\(\dfrac{4}{3}y=\sqrt{2x+1}+1+\dfrac{2}{\dfrac{3}{\sqrt{2x+1}+1}}-2\)

Áp dụng BĐT Cô-si:

\(\sqrt{2x+1}+1+\dfrac{\dfrac{2}{3}}{\sqrt{2x+1}+1}>=2\sqrt{\dfrac{2}{3}}\)

\(\Rightarrow\)\(\dfrac{4}{3}y>=2\sqrt{\dfrac{2}{3}}-2\Rightarrow y>=\dfrac{2\sqrt{\dfrac{2}{3}}-2}{4}\)

Dấu = xảy ra \(\Leftrightarrow\)x=\(\dfrac{\dfrac{2}{3}-2\sqrt{\dfrac{2}{3}}}{2}\)(có thể tính sai)

Ta xét tam giác ABC có:

\(R^2+R^2=2R^2\)

suy ra\(R^2+R^2=\left(\sqrt{2}R\right)^2\)

Suy ra

\(OA^2+OB^2=AB^2\)

\(\Rightarrow\) Tam giác ABC vuông tại O

\(\Rightarrow\)Góc AOB=90 độ

\(\Rightarrow\)Số đó cũng nhỏ AB= 90 độ( góc ở tâm)

\(\Rightarrow\)cung lớn AB=270 độ

Suy ra

\(P^2=\dfrac{x+y}{x+y-4034+4034}=\dfrac{x+y}{x+y}=1\)

Vậy P=1(vì P>0)

chết nhá sao cậu toàn hỏi bài ở lớp thế ko sợ cô mắng à

Ta có:

\(P^2\)=\(\dfrac{x+y}{x+y-4034+2\sqrt{\left(x-2017\right)\left(y-2017\right)}}\)

\(P^2\)=\(\dfrac{x+y}{x+y-4034+2\sqrt{xy-2017\left(x+y\right)+2017^2}}\)

Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2017}\)

Suy ra xy=2017(x+y)

Suy ra \(P^2=\dfrac{x+y}{x+y-4034+2\sqrt{2017\left(x+y\right)-2017\left(x+y\right)+2017^2}}\)

\(P^2=\dfrac{x+y}{x+y-4034+2\sqrt{2017^2}}\)

\(P^2=\dfrac{x+y}{x+y-4034+4034}=\dfrac{x+y}{x+y}=1\)

Vậy P=1

dùng hệ số bất định ấy ,lười lắm

Nếu x>2

suy ra \(\sqrt{5-x^2}>1\)

suy ra\(\dfrac{x^3}{\sqrt{5-x^2}}\)>2^3=8

Suy ra :\(\dfrac{x^3}{\sqrt{5-x^2}}+8x^2>32+8=40\)

Chứng minh tương tự với x<2

Với x=2 suy ra thỏa mãn

Ta có:

\(x^3=\left[\dfrac{x\left(x+1\right)}{2}\right]\)^2+\(\left[\dfrac{x\left(x+1\right)}{2}\right]^2\)(tự cm)

Thay lần lượt vào là xong

hay dùng quy nạp cũng được

Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3

2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4

...

Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)

N=\(98+\dfrac{1}{100}-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-....-\dfrac{1}{99.100}\)

Xét P=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{100}\)

Vậy N=98-1+\(\dfrac{1}{50}\)

N=\(97+\dfrac{1}{50}\)

Vậy 97<N<98(ĐPCM)