lớn hơn hay = thế ạ

Ta có :

\(a^2b+b^2c+c^2a\ge\frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)

\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)\left(1+2a^2b^2c^2\right)\ge9a^2b^2c^2\)

\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^{3v}+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)(*)

Áp dụng BĐT AM-GM ta có:

\(a^2b+a^4b^3c^2+a^3b^2c^4\ge3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)

\(b^2c+a^2b^4c^3+a^4b^3c^2\ge3a^2b^3c^2\)

\(c^2a+a^3b^2c^4+a^2b^4c^4\ge3a^2b^2c^3\)

Cộng theo vế

\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)

Vậy $(*)$ đúng

Do đó ta có đpcm

#Cừu

~

p/s: lần sau ghi nguồn 

#  https://h7.net/hoi-dap/toan-9/chung-minh-a-2b-b-2c-c-2a-9a-2b-2c-2-1-2a-2b-2c-2--faq362074.html

Đề sai nhé mọi người , đầu tiên tưởng đề đúng nhưng ko phải 

Lấy VT - VP

Phân tích ta được (a-b)(b-c)(c-a) < 0 nhé !

(a - b)(a - c)(b - c) \(\ge\)0 (đúng nhưng dấu = không xảy ra.

\(a^2b+b^2c+c^2a>ab^2+bc^2+ca^2\)

\(\Leftrightarrow a^2b+b^2c+c^2a-ab^2-bc^2-ca^2>0\)

\(\Leftrightarrow ab\left(a-b\right)+abc+bc\left(b-c\right)-abc+ca\left(c-a\right)>0\)

\(\Leftrightarrow ab\left(a-b+c\right)-bc\left(a-b+c\right)+ca\left(c-a\right)>0\)

\(\Leftrightarrow b\left(a-c\right)\left(a-b+c\right)-ca\left(a-c\right)>0\)

\(\Leftrightarrow\left(a-c\right)\left(ba-b^2+bc-ca\right)>0\)

\(\Leftrightarrow\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]>0\)

\(\Leftrightarrow\left(a-c\right)\left(a-b\right)\left(b-c\right)>0\)( đúng )

\(\Rightarrow a^2b+b^2c+c^2a>ab^2+bc^2+ca^2\)

                                                     đpcm

\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+1a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)

\(\Leftrightarrow2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2c}\right)\ge1+\dfrac{b+2a}{b+2a}+\dfrac{c+2b}{c+2b}+\dfrac{a+2c}{a+2c}=1+1+1+1=4\)Thật vậy:

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

\(\ge\dfrac{4a}{2\left(a+b+c\right)}+\dfrac{4b}{2\left(a+b+c\right)}+\dfrac{4c}{2\left(a+b+c\right)}=2\)

\(\Rightarrow VT\ge2.2=4\)

\(\RightarrowĐPCM\)

Đề bài sai

Phản ví dụ: \(a=\dfrac{1}{2};b=2;c=4\) vì VT<VP

Cho \(a=b=c\)

\(\Rightarrow2\left(\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\right)\ge1+\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\)

\(\Leftrightarrow2\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\ge1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)

\(\Leftrightarrow2\ge2\) ( Đúng)

\(\Rightarrow2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)