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BĐT Bunhiacopxky em chưa học cô ạ
Cô cong cách nào không ạ
Nguyễn Thị Nguyệt Ánh:
Vậy thì bạn có thể chứng minh $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$ thông qua BĐT Cô-si:
Áp dụng BĐT Cô-si:
$x+y+z\geq 3\sqrt[3]{xyz}$
$xy+yz+xz\geq 3\sqrt[3]{x^2y^2z^2}$
Nhân theo vế:
$(x+y+z)(xy+yz+xz)\geq 9xyz$
$\Rightarrow \frac{xy+yz+xz}{xyz}\geq \frac{9}{x+y+z}$
hay $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$
\(P=\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+2-2=\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2\)
\(=\left(\frac{a^2}{b^3}+\frac{1}{a}+\frac{1}{a}\right)+\left(\frac{b^2}{c^3}+\frac{1}{b}+\frac{1}{b}\right)+\left(\frac{c^2}{a^3}+\frac{1}{c}+\frac{1}{c}\right)-2\)
Áp dụng BĐT AM-GM cho 3 số dương:
\(\frac{a^2}{b^3}+\frac{1}{a}+\frac{1}{a}\ge3\sqrt[3]{\frac{a^2}{b^3}.\frac{1}{a}.\frac{1}{a}}=\frac{3}{b}\)
\(\frac{b^2}{c^3}+\frac{1}{b}+\frac{1}{b}\ge3\sqrt[3]{\frac{b^2}{c^3}.\frac{1}{b}.\frac{1}{b}}=\frac{3}{c}\)
\(\frac{c^2}{a^3}+\frac{1}{c}+\frac{1}{c}\ge3\sqrt[3]{\frac{c^2}{a^3}.\frac{1}{c}.\frac{1}{c}}=\frac{3}{a}\)
\(\Rightarrow P\ge\frac{3}{b}+\frac{3}{c}+\frac{3}{a}-2=3-2=1\)
Dấu "=" xảy ra khi \(a=b=c=3\)
Đặt \(\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\) thì
\(\Rightarrow\hept{\begin{cases}x+y+z=1\\P=\frac{y^3}{x^2}+\frac{z^3}{y^2}+\frac{x^3}{z^2}\end{cases}}\)
Ta có:
\(\frac{x^3}{z^2}+z+z\ge3x,\frac{y^3}{x^2}+x+x\ge3y,\frac{z^3}{y^2}+y+y\ge3z\)
\(\Rightarrow\frac{x^3}{z^2}\ge3x-2z,\frac{y^3}{x^2}\ge3y-2x,\frac{z^3}{y^2}\ge3z-2y\)
\(\Rightarrow P\ge3x-2z+3y-2x+3z-2y=x+y+z=1\)
1. Áp dụng BĐT Cauchy dạng Engle, ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
\(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)
\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)
Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)
\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)
Áp dụng BĐT Cauchy cho a ; b dương
Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có :
\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)
\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)
Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)
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Lời giải :
\(P=\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\)
\(P=\frac{1}{9}\cdot\left(\frac{9}{a+b+b}+\frac{9}{b+c+c}+\frac{9}{c+a+a}\right)\)
Áp dụng bđt Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)ta có :
\(P\le\frac{1}{9}\left(\frac{1}{a}+\frac{2}{b}+\frac{1}{b}+\frac{2}{c}+\frac{1}{c}+\frac{2}{a}\right)\)
\(=\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\frac{1}{3}\cdot9=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Theo Cauchy: \(\frac{1}{a+2b}=\frac{1}{a+b+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)\)
Tương tự hai BĐT còn lại và cộng theo vế thu được:
\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3\)
Đẳng thức xảy ra khi a = b = c = 1.
Vậy..
\(GT\Rightarrow\)\(\frac{1}{a+2}+\frac{3}{b+4}\leq1-\frac{2}{c+3}\)
Áp dụng BĐT AM-GM ta có:
\(1-\frac{2}{c+3}\geq\frac{1}{a+2}+\frac{3}{b+4}\geq2\sqrt{\frac{3}{(a+2)(b+4)}}\)
Tương tự ta có:
\(1-\frac{1}{a+2}\geq\frac{3}{b+4}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(b+4)}}\)
\(1-\frac{3}{b+4}\geq\frac{1}{a+2}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(a+2)}}\)
Nhân theo vế ta được: \((1-\frac{2}{c+3})(1-\frac{1}{a+2})(1-\frac{3}{b+4})\geq \frac{48}{(a+2)(b+4)(c+3)}\)
\(\Leftrightarrow (\frac{c+1}{c+3})(\frac{a+1}{a+2})(\frac{b+1}{b+4})\geq\frac{48}{(a+2)(b+4)(c+3)}\)
\(\Leftrightarrow(a+1)(b+1)(c+1)\geq48\)
Dấu "=" xảy ra khi \(a=1;c=3;b=5\)
\(Gt\Leftrightarrow 1-\frac{1}{a+2}+1-\frac{3}{b+4}+\frac{c+1}{c+3}\geq 2\\\Leftrightarrow \frac{a+1}{a+2}+\frac{b+1}{b+4}+\frac{c+1}{c+3}\geq 2\)
Đặt \((a+1;b+1;c+1)\rightarrow (x;y;z)\), vậy cần tìm GTNN của \(Q=xyz\)
Ta có: \(\frac{x}{x+1}+\frac{y}{y+3}+\frac{z}{z+2}\geq 2\)
Áp dụng BĐT AM-GM ta có:
\(\frac{x}{x+1}\geq 1-\frac{y}{y+3}+1-\frac{z}{z+2}=\frac{3}{y+3}+\frac{2}{z+2}\geq 2\sqrt{\frac{6}{(y+3)(z+2)}}\)
\(\frac{y}{y+3}\geq 1-\frac{x}{x+1}+1-\frac{z}{z+2}=\frac{1}{x+1}+\frac{2}{z+2}\geq 2\sqrt{\frac{2}{(x+1)(z+2)}}\)
\(\frac{z}{z+2}\geq 1-\frac{x}{x+1}+1-\frac{y}{y+3}= \frac{1}{x+1}+\frac{3}{y+3}\geq 2\sqrt{\frac{3}{(x+1)(y+3)}}\)
Nhân theo vế ta có:\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{48}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\Leftrightarrow Q\ge48\)
Dấu "=" xảy ra khi \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{x+1}=\frac{3}{y+3}=\frac{2}{z+2} & & \\ \frac{1}{a+2}+\frac{3}{b+4}=\frac{c+1}{c+3} & & \end{matrix}\right.\)\(\Leftrightarrow a=1;b=5;c=3\)