\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)

Tương tự và cộng lại;

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đề bài sai

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

Lời giải:

Đặt $a-b=x; b-c=y, c-a=z$ thì $x+y+z=0$.

ĐKĐB tương đương với:

$x^2+y^2+z^2=(y-z)^2+(z-x)^2+(x-y)^2$

$\Leftrightarrow x^2+y^2+z^2=2(x^2+y^2+z^2)-2(xy+yz+xz)$

$\Leftrightarrow x^2+y^2+z^2=2(xy+yz+xz)$
$\Leftrightarrow 2(x^2+y^2+z^2)=x^2+y^2+z^2+2(xy+yz+xz)$

$\Leftrightarrow 2(x^2+y^2+z^2)=(x+y+z)^2=0$

$\Rightarrow x=y=z=0$

$\Leftrightarrow a-b=b-c=c-a=0$

$\Leftrightarrow a=b=c$ (ta có đpcm)

Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\) thì ta có \(x+y+z=0\). Điều kiện đã cho tương đương \(x^2+y^2+z^2=\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2=2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2=4\left(xy+yz+zx\right)\)

\(\Leftrightarrow4\left(xy+yz+zx\right)=0\)

\(\Leftrightarrow xy+yz+zx=0\)

\(\Leftrightarrow x^2+y^2+z^2=0\)

\(\Leftrightarrow x=y=z=0\)

\(\Leftrightarrow a-b=b-c=c-a=0\)

\(\Leftrightarrow a=b=c\)

Ta có đpcm

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

(a-b)^2 + (b-c)^2 + (c-a)^2 = (a+b-2c)^2 + (b+c-2a)^2 + (c+a-2b)^2

<=> (a+b-2c)^2 - (a-b)^2 + (b+c-2a)^2 - (b-c)^2 + (c+a-2b)^2 - (c-a)^2 = 0

<=> (2b-2c)(2a-2c) + (2c-2a)(2b-2a) + (2a-2b)(2c-2b) = 0

<=> (b-c)(a-c) + (c-a)(b-a) + (a-b)(c-b) = 0

<=> ab - ac - bc + c^2 + bc - ab - ac - a^2 + ac - bc - ab + b^2 = 0

<=> a^2 + b^2 + c^2 - ab - bc - ac = 0

<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0

<=> (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) = 0

<=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0

<=> (a-b)^2=0; (b-c)^2=0; (c-a)^2=0

<=> a-b=0; b-c=0; c-a=0

<=> a=b=c (đpcm)