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Bài 1:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)
\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ab-ac}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)
\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Bài 2:
\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))
\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)
chị giải thích cho em cái đoạn này với ạ
\(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)
\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)
Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)
\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
Áp dụng BĐT Cauchy ta có
\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh
\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)
Áp dụng BĐT Cauchy:
\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm
Áp dụng BĐT Cauchy Swarch
\(\Sigma\dfrac{1}{a^2+2bc}\ge\dfrac{9}{\left(a+b+c\right)^2}=9\)
Vậy Min ... =9 khi a=b=c=1/3
Áp dụng BĐT Cauchy - Schwarz vào bài toán , ta có :
\(Q=\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\dfrac{9}{\left(a+b+c\right)^2}=\dfrac{9}{1^2}=9\) Dấu " = " xảy ra khi : \(\dfrac{1}{a^2+2ab}=\dfrac{1}{b^2+2ac}=\dfrac{1}{c^2+2ab}\Leftrightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow Q_{Min}=9\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz và AM-GM:
$M=\frac{b^2+c^2}{a^2}+a^2(\frac{1}{b^2}+\frac{1}{c^2})$
$\geq \frac{b^2+c^2}{a^2}+a^2.\frac{4}{b^2+c^2}$
$=(\frac{b^2+c^2}{a^2}+\frac{a^2}{b^2+c^2})+\frac{3a^2}{b^2+c^2}$
$\geq \sqrt{\frac{b^2+c^2}{a^2}.\frac{a^2}{b^2+c^2}}+\frac{3(b^2+c^2)}{b^2+c^2}$
$=2+3=5$
Vậy $M_{\min}=5$
\(VT=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\)
\(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
từ (a+b+c)^2=a^2+b^2+c^2
suy ra ab+bc+ac=0suy ra ab=-(bc+ac);ac=-(ab+bc);bc=-(ab+ac)
xét a^2+2bc=a^2+bc-ab-ac=(a-c)(a-b)
tương tự dc b^2+2ac=(b-a)(b-c)
c^2+2ab=(a-c)(b-c)
thay vao điều phải c/m dc
a^2/(a-c)(a-b) -b^2/(a-b)(b-c) +c^2(a-c)(b-c)
=a^2b-a^2c-b^2a+b^2c+c^2a-bc^2/(a-b)(a-c)(b-c)
=abc(ac-bc+bc-ab+ab-ac)/(a-b)(a-c)(b-c)=0