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BĐT cần chứng minh tương đương :
\(\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\ge\dfrac{ab+bc+ac}{abc}\)
\(\Leftrightarrow\dfrac{a^8+b^8+c^8}{a^2b^2c^2}\ge ab+bc+ac\)
\(\Leftrightarrow\dfrac{a^6}{b^2c^2}+\dfrac{b^6}{a^2c^2}+\dfrac{c^6}{a^2b^2}\ge ab+bc+ac\)
Do \(a^2+b^2+c^2\ge ab+bc+ac\)
Ta phải cm
\(\dfrac{a^6}{b^2c^2}+\dfrac{b^6}{a^2c^2}+\dfrac{c^6}{a^2b^2}\ge a^2+b^2+c^2\)(1)
Đặt : \(\left(a^2;b^2;c^2\right)=\left(x;y;z\right)\)
\(\Rightarrow\left(1\right)\Leftrightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}\ge x+y+z\)
Áp dụng C.B.S
\(\Rightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}=\dfrac{x^4}{xyz}+\dfrac{y^4}{xyz}+\dfrac{z^4}{xyz}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3xyz}\)
Theo Bunhiacopxki: \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\)\(\Rightarrow\left(x^2+y^2+z^2\right)^2\ge\dfrac{\left(x+y+z\right)^4}{9}\)
Theo Cauchy : \(\Rightarrow3xyz\le\dfrac{\left(x+y+z\right)^3}{9}\)
\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3xyz}\ge\dfrac{\dfrac{\left(x+y+z\right)^4}{9}}{\dfrac{\left(x+y+z\right)^3}{9}}=x+y+z\)
\(\Rightarrow\)\(\Rightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}\ge x+y+z\)
=> đpcm
BĐT cần chứng minh tương đương :
a8+b8+c8a3b3c3≥ab+bc+acabca8+b8+c8a3b3c3≥ab+bc+acabc
⇔a8+b8+c8a2b2c2≥ab+bc+ac⇔a8+b8+c8a2b2c2≥ab+bc+ac
⇔a6b2c2+b6a2c2+c6a2b2≥ab+bc+ac⇔a6b2c2+b6a2c2+c6a2b2≥ab+bc+ac
Do a2+b2+c2≥ab+bc+aca2+b2+c2≥ab+bc+ac
Ta phải cm
a6b2c2+b6a2c2+c6a2b2≥a2+b2+c2a6b2c2+b6a2c2+c6a2b2≥a2+b2+c2(1)
Đặt : (a2;b2;c2)=(x;y;z)(a2;b2;c2)=(x;y;z)
⇒(1)⇔x3yz+y3xz+z3xy≥x+y+z⇒(1)⇔x3yz+y3xz+z3xy≥x+y+z
Áp dụng C.B.S
⇒x3yz+y3xz+z3xy=x4xyz+y4xyz+z4xyz≥(x2+y2+z2)23xyz⇒x3yz+y3xz+z3xy=x4xyz+y4xyz+z4xyz≥(x2+y2+z2)23xyz
Theo Bunhiacopxki: x2+y2+z2≥(x+y+z)23x2+y2+z2≥(x+y+z)23⇒(x2+y2+z2)2≥(x+y+z)49⇒(x2+y2+z2)2≥(x+y+z)49
Theo Cauchy : ⇒3xyz≤(x+y+z)39⇒3xyz≤(x+y+z)39
⇒(x2+y2+z2)23xyz≥(x+y+z)49(x+y+z)39=x+y+z⇒(x2+y2+z2)23xyz≥(x+y+z)49(x+y+z)39=x+y+z
⇒⇒⇒x3yz+y3xz+z3xy≥x+y+z⇒x3yz+y3xz+z3xy≥x+y+z
=> đpcm
Áp dụng BĐT: x2+y2+z2\(\ge\)xy+yz+zx ( với x,y,z >0)
Ta có\(\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\)\(\ge\)\(\dfrac{a^4b^4+b^4c^4+c^4a^4}{a^3b^3c^3}\)
\(\ge\)\(\dfrac{a^4b^2c^2+b^4c^2a^2+c^4a^2b^2}{a^3b^3c^3}\)=\(\dfrac{a^2+b^2+c^2}{abc}\)\(\ge\)\(\dfrac{ab+bc+ca}{abc}\)
= \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (đpcm)
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c
b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)
Tương tự cho 2 cái kia rồi cộng lại
\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)
\(VT=\dfrac{a^3bc}{c+ab^2c}+\dfrac{ab^3c}{a+abc^2}+\dfrac{abc^3}{b+a^2bc}\)
\(=abc\left(\dfrac{a^2}{c+ab^2c}+\dfrac{b^2}{a+abc^2}+\dfrac{c^2}{b+a^2bc}\right)\)
Áp dụng bđt Cauchy-Schwarz dạng engel có:
\(VT\ge\dfrac{abc\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}\)\(=\dfrac{abc\left(a+b+c\right)}{1+abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
Vậy...
Sai đề không bạn,tại a=b=c=2 thay vào không thỏa mãn nha
Em kiểm tra lại đề bài
\(4a^2+b^2+3c^2=4ab\Leftrightarrow\left(2a-b\right)^2+3c^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a-b=0\\c=0\end{matrix}\right.\)
Dẫn tới biểu thức P không xác định
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{b}{2}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{bc+ab}{c+a}+\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{6}\)
Dấu "=" xảy ra khi \(a=b=c\)
Lời giải:
Ta có:
\(\frac{a^8+b^8+c^8}{a^3b^3c^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow a^8+b^8+c^8\geq a^2b^2c^2(ab+bc+ac)(*)\)
Áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} a^8+b^8\geq 2a^4b^4\\ b^8+c^8\geq 2b^4c^4\\ c^8+a^8\geq 2c^4a^4\end{matrix}\right.\Rightarrow a^8+b^8+c^8\geq a^4b^4+b^4c^4+c^4a^4\)
Tiếp tục áp dụng AM-GM:
\(a^8+b^8+a^4b^4+c^8\geq 4\sqrt[4]{a^{12}b^{12}c^8}=4a^3b^3c^2\)
\(b^8+c^8+b^4c^4+a^8\geq 4b^3c^3a^2\)
\(c^8+a^8+c^4a^4+b^8\geq 4c^3a^3b^2\)
Cộng lại: \(3(a^8+b^8+c^8)+(a^4b^4+b^4c^4+c^4a^4)\geq 4a^2b^2c^2(ab+bc+ca)\)
Mà \(a^8+b^8+c^8\geq a^4b^4+b^4c^4+c^4a^4\Rightarrow 4(a^8+b^8+c^8)\geq 4a^2b^2c^2(ab+bc+ac)\)
hay \(a^8+b^8+c^8\geq a^2b^2c^2(ab+bc+ac)\Rightarrow (*)\) đúng
Ta có đpcm.