Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)
Đặt \(\left\{{}\begin{matrix}x=a+b+c\\y=ab+bc+ca\end{matrix}\right.\) khi đó \(BDT\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}\le\dfrac{12+4x+y}{9+4x+2y}\)
\(\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}-1\le\dfrac{12+4x+y}{9+4x+2y}-1\)
\(\Leftrightarrow\dfrac{2x+3-xy}{x^2+2x+y+xy}\le\dfrac{3-y}{9+4x+2y}\)
\(\Leftrightarrow\dfrac{5x^2-3x^2y-xy^2-6xy+24x+y^2+3y+27}{\left(4x+2y+9\right)\left(x^2+xy+2x+y\right)}\le0\)
Đúng vì \(\dfrac{5}{3}x^2y\ge5x^2;\dfrac{x^2y}{3}\ge y^2;xy^2\ge9x;5xy\ge15x;xy\ge3y;x^2y\ge27\)
\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)
\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Với a, b, c > 0 có:
\(P=\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\\ =\dfrac{a^2}{a\left(b+2c\right)}+\dfrac{b^2}{b\left(c+2a\right)}+\dfrac{c^2}{c\left(a+2b\right)}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\)
chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\)
Bài 2:
Áp dụng BĐT: \(x^2+y^2+z^2\ge xy+yz+xz\), ta có:
\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2\) (1)
Lại áp dụng tương tự ta có:
\(\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\ge ab^2c+abc^2+a^2bc\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\) (2)
Từ (1) và (2) suy ra:
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Bài 1:
Áp dụng BĐT Cô -si, ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\sqrt[3]{\dfrac{b^2}{c^3}.\dfrac{1}{b}.\dfrac{1}{b}}=\dfrac{3}{c}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng vế theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
p/s: không chắc lắm, có gì sai xót xin giúp đỡ
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Leftrightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Áp dụng BĐT AM-GM:
\(\frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}(\frac{b}{a+b}+\frac{c}{a+c})\)
Hoàn toàn tương tự:
\(\frac{1}{\sqrt{b^2+1}}=\sqrt{\frac{ac}{(b+a)(b+c)}}\leq \frac{1}{2}(\frac{a}{b+a}+\frac{c}{b+c})\)
\(\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}(\frac{a}{c+a}+\frac{b}{b+c})\)
Cộng theo vế:
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\leq \frac{1}{2}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
\(BDT\Leftrightarrow\dfrac{\dfrac{1}{a}+\dfrac{1}{a^2}}{1+\dfrac{1}{a}+\dfrac{1}{a^2}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{b^2}}{1+\dfrac{1}{b}+\dfrac{1}{b^2}}+\dfrac{\dfrac{1}{c}+\dfrac{1}{c^2}}{1+\dfrac{1}{c}+\dfrac{1}{c^2}}\le2\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(n,h,t\right)\) thì ta có :
\(\Leftrightarrow\dfrac{n+n^2}{1+n+n^2}+\dfrac{h+h^2}{1+h+h^2}+\dfrac{t+t^2}{1+t+t^2}\le2\)
\(\Leftrightarrow\dfrac{1}{1+n+n^2}+\dfrac{1}{1+h+h^2}+\dfrac{1}{1+t+t^2}\ge1\)
Đặt \(n=\dfrac{yz}{x^2};h=\dfrac{xz}{y^2};t=\dfrac{xy}{z^2}\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\)
Và \(\dfrac{x^4}{x^4+x^2yz+y^2z^2}+\dfrac{y^4}{y^4+xy^2z+x^2z^2}+\dfrac{z^4}{z^4+xyz^2+x^2y^2}\ge1\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2}\)
Cần cm \(\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2}\ge1\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\ge x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2\)
\(\Leftrightarrow x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2yz+xy^2z+xyz^2\left(1\right)\)
Áp dụng BĐT AM-GM ta có:
\(x^2y^2+y^2z^2=y^2\left(x^2+z^2\right)\ge2xy^2z\)
Tương tự rồi cộng theo vế ta có \(\left(1\right)\) đúng
Khi \(a=b=c=1\)
Sửa đề\(VP\le 2\) sau đó nó chính là 1 dạng của BĐT Vasc k cần thêm j cả :">
Ta đi chứng minh BĐT : \(a^2+b^2+c^2\ge2\left(bc+ac-ab\right)\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\) \(\left(a+b-c\right)^2\ge0\) luôn đúng.
\(\Rightarrow2\left(bc+ac-ab\right)\le\dfrac{5}{3}\)
\(\Leftrightarrow bc+ac-ab\le\dfrac{5}{6}< 1\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}< \dfrac{1}{abc}\)