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a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
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cho 2 biểu thức mà c/m 1 biểu thức M là sao
Biểu thức N vứt sọt à hay làm cái j v :V
tớ cũng nghĩ vậy nhưng mãi sau mới biết chứng minh M =N rồi chứng minh N >=(a+b+c)/8 để suy ra M >=(a+b+c)/8
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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
\(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
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1/ Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel :
\(A\ge\frac{\left(a+b+c\right)^2}{3\left(a+b+c\right)}=\frac{a+b+c}{3}=\frac{3}{3}=1\)
Dấu "=" xảy ra <=> a=b=c=1
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Áp dụng bđt AM-GM:
\(M\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}+\dfrac{b^3}{b^2+\dfrac{b^2+c^2}{2}+c^2}+\dfrac{c^3}{c^2+\dfrac{a^2+c^2}{2}+a^2}\)
\(=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}+\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}+\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\)
\(=\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)
Xét:
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\)
\(=a-\dfrac{ab^2}{a^2+b^2}+b-\dfrac{b^2c}{b^2+c^2}+c-\dfrac{c^2a}{c^2+a^2}\)
\(\ge a+b+c-\dfrac{ab^2}{2ab}-\dfrac{b^2c}{2bc}-\dfrac{c^2a}{2ac}=a+b+c-\dfrac{a}{2}-\dfrac{b}{2}-\dfrac{c}{2}=\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow M\ge1."="\Leftrightarrow a=b=c=1\)
dòng thứ 5 từ dưới lên cái đầu là bc^2 nhé. Cái sau là ca^2
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\(\dfrac{a^3}{a^2+bc}=a-\dfrac{abc}{a^2+bc}\ge a-\dfrac{abc}{2a\sqrt{bc}}=a-\dfrac{\sqrt{bc}}{2}\)
\(\dfrac{b^3}{b^2+ca}\ge b-\dfrac{\sqrt{ac}}{2};\dfrac{c^3}{c^2+ab}\ge c-\dfrac{\sqrt{ab}}{2}\)
\(\Rightarrow M\ge a+b+c-\left(\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{bc}}{2}+\dfrac{\sqrt{ca}}{2}\right)=2022-\left(\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\right)\)
\(do:\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)
\(\Rightarrow M\ge2022-\dfrac{a+b+c}{2}=2022-\dfrac{2022}{2}=1011\)
\(min_M=2021\Leftrightarrow a=b=c=674\)
có đoạn bạn sửa lại tí nhé tại lúc đầu mình đọc đề thành \(a+b+c=2022\)
\(M\ge a+b+c-\left(\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\right)\ge a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\ge\dfrac{2022}{2}=1011\)
c/m phản chứng hoặc đổi (a^3;b^3)->(x;y) ,dùng holder
AM-GM: \(a^3+b^3\ge\frac{\left(a+b\right)^3}{4}\)