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\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)
\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
P=\(\dfrac{\sqrt{2}.a}{\sqrt{\left(a^2+\left(b+c\right)^2\right)\left(1+1\right)}}+\dfrac{\sqrt{2}.b}{\sqrt{\left(b^2+\left(a+c\right)^2\right)\left(1+1\right)}}+\dfrac{\sqrt{2}.c}{\sqrt{\left(c^2+\left(b+a\right)^2\right)\left(1+1\right)}}\)>=\(\dfrac{\sqrt{2}.a}{\sqrt{\left(a+b+c\right)^2}}+\dfrac{\sqrt{2}.b}{\sqrt{\left(a+b+c\right)^2}}+\dfrac{\sqrt{2}.c}{\sqrt{\left(a+b+c\right)^2}}\)>=\(\sqrt{2}\)
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
\(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
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
Ta có :\(\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}=\dfrac{1}{\sqrt{\left(4a^2+4ab+b^2\right)+\left(a^2-2ab+b^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2a+b\right)^2+\left(a-b\right)^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{2a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\) (Cosi)
Tương tự cộng lại ta được :
\(P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}=\dfrac{1}{\sqrt{3}}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\sqrt{3}\)
\(\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)\(\le\) \(\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}\) làm thế nào hả bn ?
Theo giả thiết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Leftrightarrow\) \(\dfrac{ab+bc+ca}{abc}=0\)
\(\Leftrightarrow\) ab+bc+ca = 0 \(\Rightarrow\) \(\left\{{}\begin{matrix}ab=-ac-bc\\bc=-ab-ac\\ac=-bc-ab\end{matrix}\right.\)
Xét \(\dfrac{a^2}{a^2+2bc}\) = \(\dfrac{a^2}{a^2+bc-ab-ac}\) = \(\dfrac{a^2}{a\left(a-c\right)-b\left(a-c\right)}\)
= \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)
CMTT \(\dfrac{b^2}{b^2+2ac}\) = \(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)
\(\dfrac{c^2}{c^2+2ab}\) = \(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow\) p = \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)+\(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)+\(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
= \(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\) = 1