\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)

Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)

Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)

\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)

\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)

\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)

\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)

\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)

\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)

\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)

\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)

Từ giả thiết \(2\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{2}{3}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)

Đặt \(Q=\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}\)

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

\(\ge\dfrac{a^2b^2c^2+3\sqrt[3]{\left(a^2b^2c^2\right)^2}+3\sqrt[3]{a^2b^2c^2}+1}{abc}=\dfrac{\left(\sqrt[3]{a^2b^2c^2}+1\right)^3}{abc}\)

\(=\left(\dfrac{\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3=\left(\sqrt[3]{abc}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)

\(=\left(\sqrt[3]{abc}+\dfrac{4}{9\sqrt[3]{abc}}+\dfrac{5}{9\sqrt[3]{abc}}\right)^3\ge\left(2\sqrt[]{\dfrac{4\sqrt[3]{abc}}{9\sqrt[3]{abc}}}+\dfrac{5}{9.\dfrac{2}{3}}\right)^3=\dfrac{2197}{216}\)

\(\Rightarrow P\ge3\sqrt[3]{\dfrac{2197}{216}}=\dfrac{13}{2}\)

Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)

\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)

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

\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)

\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).

ĐTXR \(\Leftrightarrow a=b=c=1\)

\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D 

oh no bài thứ nhất là dạng chứng minh cs đúng ko ,

ko thể nào là dạng tìm a,b,c đc-.-

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2022}\)

hay \(\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+2abc=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

-Xét a + b = 0 => P = 2022^2021

Bạn xét tương tự với b + c = 0 và c + a = 0 dc P = 2022^2021 nhé

a+bab+a+bc(a+b+c)=0a+bab+a+bc(a+b+c)=0

(a+b)[ab+bc+ca+c2abc(a+b+c)]=0(a+b)[ab+bc+ca+c2abc(a+b+c)]=0

 ⇔ a=−b

 ⇔ b=−c

 ⇔ c=−a

Thay vào P từng cái rồi tính tiếp nhé

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) và BĐT cần chứng minh là:

\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel và AM-GM ta có:

\(VT=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}=VP\)

Xảy ra khi \(x=y=z=1 \Rightarrow a=b=c=1\)

ai tick cho mik , mik tick lại cho !^__<nhớ giải câu hỏi nhé ! thanks