Bài này mẫu số là \(\left(a+b+c\right)^3\) thì đúng hơn, mũ 2 cách làm vẫn y hệt nhưng cho 1 kết quả rất xấu

\(A\ge3\left(a^2+b^2+c^2\right)+\dfrac{24\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(=3\left(a+b+c\right)^2+\dfrac{192}{a+b+c}-48\)

\(=\dfrac{\sqrt{6}}{3}\left(a+b+c\right)^2+\dfrac{96}{a+b+c}+\dfrac{96}{a+b+c}+\left(3-\dfrac{\sqrt{6}}{3}\right)\left(a+b+c\right)^2-48\)

\(\ge3\sqrt[3]{\dfrac{96^2.\sqrt{6}}{3}}+\left(3-\dfrac{\sqrt{6}}{3}\right).3\left(ab+bc+ca\right)-48=...\)

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

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

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

Do đó ta chỉ cần chứng minh \(\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge2\)

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

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

\(\Leftrightarrow3abc\ge4\left(ab+bc+ca\right)-9\)

\(\Rightarrow\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge\dfrac{a^2+b^2+c^2+4\left(ab+bc+ca\right)-9}{ab+bc+ca}\)

\(=\dfrac{\left(a+b+c\right)^2-9+2\left(ab+bc+ca\right)}{ab+bc+ca}=2\) (đpcm)

sai cơ bản rồi bạn ơi : a(a+bc)^2 không bằng dc (a^2+abc)^2

a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)

b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)

\(=\left(a-b\right)\left(a-c\right)\)

Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)

Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)

Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)

\(\left(c;d\right)\Rightarrow\left(-c;-d\right)\)

\(\left(a-1\right)^2+\left(b-1\right)^2=1\)

\(\left(c-5\right)^2+\left(d-5\right)^2=100\)

Gọi \(A\left(a;b\right)\) thuộc đường tròn có pt \(\left(x-1\right)^2+\left(y-1\right)^2=1\) (C) có tâm \(I\left(1;1\right)\) bán kính \(R=1\)

\(B\left(d;c\right)\) thuộc đường tròn có pt \(\left(x-5\right)^2+\left(y-5\right)^2=100\) (C') có tâm \(I'\left(5;5\right)\) bán kính \(R=10\)

\(\Rightarrow AB^2=P=\left(a-d\right)^2+\left(b-c\right)^2\)

\(P_{min}\Leftrightarrow A;B\) là giao điểm nằm cùng phía so với I và I' của đường thẳng II' với 2 đường tròn

Phương trình II': \(x-y=0\)

\(\Rightarrow A\left(\dfrac{2-\sqrt{2}}{2};\dfrac{2-\sqrt{2}}{2}\right)\) ; \(B\left(5-5\sqrt{2};5-5\sqrt{2}\right)\)

\(\Rightarrow P_{min}=AB=\dfrac{9\sqrt{2}-8}{\sqrt{2}}=9-4\sqrt{2}\)

\(a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+c^2+2bc\Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự : \(b^2-a^2-c^2=2ac\) ; \(c^2-a^2-b^2=2ab\)

Ta có : \(T=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{a^2}{2bc}+\frac{b^2}{2ca}+\frac{c^2}{2ab}\)

\(=\frac{1}{2abc}\left(a^3+b^3+c^3\right)\)(1)

Ta sẽ chứng minh nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\)

Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

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

= 0

=> \(a^3+b^3+c^3=3abc\) thay vào (1) được : 

\(T=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)

Ta có:

 \(\frac{\left(1+a\right)^2\left(1+b\right)^2}{1+c^2}=\frac{\left(1+a+b+ab\right)^2}{1+c^2}\)

\(\ge\frac{4\left(a+b\right)\left(1+ab\right)}{1+c^2}=\frac{4a+4ab^2+4b+4a^2b}{1+c^2}\)

\(=4a\frac{1+b^2}{1+c^2}+4b\frac{1+a^2}{1+c^2}\)

Tương tự : 

\(\frac{\left(1+b\right)^2\left(1+c\right)^2}{1+a^2}\ge4c\frac{1+b^2}{1+a^2}+4b\frac{1+c^2}{1+a^2}\)

\(\frac{\left(1+c\right)^2\left(1+a\right)^2}{1+b^2}\ge4a\frac{1+c^2}{1+b^2}+4c\frac{1+a^2}{1+b^2}\)

Đến đây dùng Cauchy là ra

Dấu = xảy ra khi a=b=c=1

bài này hay đó

được đó nhok, thanks nhok nhiều ^_^

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

Áp dụng BĐT C-S:

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

Đặt \(a^2+b^2+c^2=x\)

Ta cần c/m: \(\dfrac{x+4}{3-x}\ge6\Leftrightarrow x+4\ge18-6x\)

\(\Leftrightarrow x\ge2\) (đúng)

Dấu = xảy ra khi \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)