B1:

\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

Xét hiệu:

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\)

\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)

\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)

=> BĐT luôn đúng

*

Ta có:

\(a< b+c\Rightarrow a^2< ab+ac\)

\(b< a+c\Rightarrow b^2< ab+ac\)

\(c< a+b\Rightarrow a^2< ac+bc\)

Cộng từng vế bất đẳng thức ta được:

\(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

Vậy: \(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

B2:

Ta có: \(a+b>c\) ; \(b+c>a\); \(a+c>b\)

Xét:\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)

\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{a+b+c}+\dfrac{1}{a+c+b}=\dfrac{2}{a+b+c}>\dfrac{2}{b+c+b+c}=\dfrac{1}{b+c}\)

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+c+a+c}=\dfrac{1}{a+c}\)

Suy ra:

\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b}\)

\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)

=> ĐPCM

@Nguyễn Thanh Hằng đọc xong xóa đii nha

\(=\left(a+b-c\right)\left(a-b\right)^2\) nha ! 

P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.

Giỏi quá à :3

Trước hết theo BĐT Schur bậc 3 ta có:

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)^2}{a^2c^2+2ab^2c}\)

\(\Rightarrow P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)

Áp dụng (1):

\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Anh giúp em câu này ạ, câu này hơi khó anh ạ, làm chắc cũng lâu, có gì anh để mai cũng được ạ! 

https://hoc24.vn/cau-hoi/cho-hinh-chop-sabcd-co-day-la-hinh-binh-hanh-m-va-p-la-hai-diem-lan-luot-di-dong-tren-ad-va-sc-sao-cho-mamd-pspc-x-x0-mat-phang-a-di-qua-m-va-song-song-voi-sab-cat-hinh-chop-sabcd-t.8753881358034

\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)

Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)

\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)

Cộng vế:

\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)

\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

chết đăng nhầm sogy nha