Ôn tập cuối năm phần số học

Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\left(a-b\right)^2\ge0\) ta có:

\(\dfrac{1}{A+B-C}+\dfrac{1}{B+C-A}\ge\dfrac{4}{A+B-C+B+C-A}=\dfrac{4}{2B}=\dfrac{2}{B}\)

\(\dfrac{1}{B+C-A}+\dfrac{1}{C+A-B}\ge\dfrac{4}{B+C-A+C+A-B}=\dfrac{4}{2C}=\dfrac{2}{C}\)

\(\dfrac{1}{C+A-B}+\dfrac{1}{A+B-C}\ge\dfrac{4}{C+A-B+A+B-C}=\dfrac{4}{2A}=\dfrac{2}{A}\)

Cộng theo vế 3 BĐT trên ta có:

\(2VT\ge\dfrac{2}{A}+\dfrac{2}{B}+\dfrac{2}{C}=2\left(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right)=2VP\Leftrightarrow VT\ge VP\)

Mình nghĩ là nó áp dụng vào bđt côsi

Xét \(a+b=0\) thì ta có ĐPCM

Xét \(b=c\)

\(\Rightarrow a=2c\)

Ta chứng minh:

\(\dfrac{8c^3+c^3}{8c^3+c}=\dfrac{2c+c}{2c+c}\)

\(\Leftrightarrow1=1\) đúng

Xét \(\left\{{}\begin{matrix}a+b\ne0\\b\ne c\end{matrix}\right.\)

Ta chứng minh:

\(\dfrac{a^3+b^3}{a^3+c^3}=\dfrac{a+b}{a+c}\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}=\dfrac{a+b}{a+c}\)

\(\Leftrightarrow\dfrac{a^2-ab+b^2}{a^2-ac+c^2}=1\)

\(\Leftrightarrow a^2-ab+b^2=a^2-ac+c^2\)

\(\Leftrightarrow a\left(c-b\right)=\left(c-b\right)\left(c+b\right)\)

\(\Leftrightarrow a=c+b\) đúng

Vậy ta có ĐPCM

Ta có: \(\left\{{}\begin{matrix}a_1^2+a_2^2\ge2a_1a_2\\a_1^2+a_3^2\ge2a_1a_3\\...................\\a_{n-1}^2+a_n^2\ge2a_{n-1}a_n\end{matrix}\right.\)

\(\Rightarrow\left(n-1\right)\left(a_1^2+a_2^2+...+a_n^2\right)\ge2\left(a_1a_2+a_1a_3+...+a_{n-1}a_n\right)\)

\(\Leftrightarrow n\left(a_1^2+a_2^2+...+a_n^2\right)\ge2\left(a_1a_2+a_1a_3+...+a_{n-1}a_n\right)+\left(a_1^2+a_2^2+...+a_n^2\right)\)

\(\Leftrightarrow n\left(a_1^2+a_2^2+...+a_n^2\right)\ge\left(a_1+a_2+...+a_n\right)^2\)

Áp dụng BĐT căn trung bình bình phương ta có:

\(\sqrt{\dfrac{a_1^2+a_2^2+....+a^2_n}{n}}\ge\dfrac{a_1+a_2+...+a_n}{n}\)

\(\Leftrightarrow\dfrac{a_1^2+a_2^2+....+a^2_n}{n}\ge\left(\dfrac{a_1+a_2+...+a_n}{n}\right)^2\)

\(\Leftrightarrow\dfrac{a_1^2+a_2^2+....+a^2_n}{n}\ge\dfrac{\left(a_1+a_2+...+a_n\right)^2}{n^2}\)

\(\Leftrightarrow a_1^2+a_2^2+....+a^2_n\ge\dfrac{\left(a_1+a_2+...+a_n\right)^2}{n}\)

\(\Leftrightarrow n\left(a_1^2+a_2^2+....+a^2_n\right)\ge\left(a_1+a_2+...+a_n\right)^2\)

Khi \(a_1=a_2=...=a_n\)

BĐT Bunyakovsky

Ta có:

\(\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)=2006\)

Dễ thấy \(\left\{{}\begin{matrix}\sqrt{a^2+2006}-a\ne0\\\sqrt{b^2+2006}-b\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+\sqrt{a^2+2006}\right)\left(\sqrt{a^2+2006}-a\right)\left(b+\sqrt{b^2+2006}\right)=2006\left(\sqrt{a^2+2006}-a\right)\\\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)\left(\sqrt{b^2+2006}-b\right)=2006\left(\sqrt{b^2+2006}-b\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2006\left(b+\sqrt{b^2+2006}\right)=2006\left(\sqrt{a^2+2006}-a\right)\\2006\left(a+\sqrt{a^2+2006}\right)=2006\left(\sqrt{b^2+2006}-b\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+\sqrt{b^2+2006}=\sqrt{a^2+2006}-a\\a+\sqrt{a^2+2006}=\sqrt{b^2+2006}-b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=\sqrt{a^2+2006}-\sqrt{b^2+2006}\left(1\right)\\a+b=\sqrt{b^2+2006}-\sqrt{a^2+2006}\left(2\right)\end{matrix}\right.\)

Lấy (1) + (2) ta được

\(a+b=0\)

Ta có : \(\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)=2006\) (*)

Nhân liên hợp ta được :

(*)\(\Leftrightarrow\dfrac{\left(a+\sqrt{a^2+2006}\right)\left(a-\sqrt{a^2+2006}\right)}{a-\sqrt{a^2+2006}}.\)\(\dfrac{\left(b+\sqrt{b^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}{b-\sqrt{b^2-2006}}=2006\)

\(\Leftrightarrow\dfrac{a^2-a^2-2006}{a-\sqrt{a^2+2006}}.\dfrac{b^2-b-2006}{b-\sqrt{b^2+2006}}=2006\)

\(\Leftrightarrow\left(-2006\right).\left(-2006\right)\dfrac{1}{\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}=2006\)

\(\Leftrightarrow\)\(\Leftrightarrow\dfrac{1}{\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}=\dfrac{1}{2006}\)

=> \(\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)=2006\) (**)

Từ (*) và (**) ta suy ra :

\(\dfrac{\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}{\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)}=1\)

Và \(\dfrac{a-\sqrt{a^2+2006}}{a+\sqrt{a^2+2006}}=\dfrac{b+\sqrt{b^2+2006}}{b-\sqrt{b^2+2006}}\)

=> \(\dfrac{a-\sqrt{a^2+2006}}{a+\sqrt{a^2+2006}}=\dfrac{b+\sqrt{b^2+2006}}{b-\sqrt{b^2+2006}}=\dfrac{1}{2}\)

+ , \(\dfrac{a-\sqrt{a^2+2006}}{a+\sqrt{a^2+2006}}=\dfrac{1}{2}\Rightarrow2a-2\sqrt{a^2+2006}=a+\sqrt{a^2+2006}\Rightarrow a=3\sqrt{a^2+2006}\)

Tương tự : b = \(3\sqrt{b^2+2006}\)

=> a+b = \(3\left(\sqrt{a^2+2006}+\sqrt{b^2+2006}\right)\)

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không biết hướng làm này có đúng không nữa ... tại còn dính ẩn ...

Ta có:

\(a^2=\left(-b-c\right)^2\)

\(\Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự ta cũng có

\(\left\{{}\begin{matrix}b^2-c^2-a^2=2ca\\c^2-a^2-b^2=2ab\end{matrix}\right.\)

Thế vô ta được

\(A=\sqrt{\dfrac{3a^2}{bc}+\dfrac{3b^2}{ca}+\dfrac{3c^2}{ab}}\)

\(=\sqrt{\dfrac{3\left(a^3+b^3+c^3\right)}{abc}}\)

\(=\sqrt{3.\dfrac{\left(a^3+b^3+c^3-3abc\right)+3abC}{abc}}\)

\(=\sqrt{3.\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{abc}}\)

\(=\sqrt{3.3}=3\)

ĐPCM

\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

Ta có :\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}\) Lại có: a³+b³+c³= [a³+3ab(a+b)+b³] +c³ - 3ab(a+b) =(a+b)³+c³-3ab(a+b) = (a+b+c)[a²+2ab+b²-ac-bc+c²] -3ab(a+b) = -3ab(a+b) = -3ab(-c) = 3abc => A=\(\dfrac{3abc}{abc}=3\)

\(P=\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)

Áp dụng BĐT Cô-si vào 3 số dương ta có :

\(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b}{3}+\dfrac{2c+a}{9}\ge3\sqrt[3]{\dfrac{a^3}{b\left(2c+a\right)}.\dfrac{b}{3}.\dfrac{2c+a}{9}}=a\) ( 1 )

Tương tự ta có :

\(\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c}{3}+\dfrac{2a+b}{9}\ge3\sqrt[3]{\dfrac{b^3}{c\left(2a+b\right)}.\dfrac{c}{3}.\dfrac{2a+b}{9}}=b\) ( 2 )

\(\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{a}{3}+\dfrac{2b+c}{9}\ge3\sqrt[3]{\dfrac{c^3}{a\left(2b+c\right)}.\dfrac{a}{3}.\dfrac{2b+c}{9}}=c\) ( 3 )

Cộng từng vế của ( 1 ) ( 2 ) và ( 3 ) ta có :

\(\dfrac{a^3}{c\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{2}{3}\left(a+b+c\right)\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{2}{3}.3\ge3\)

\(\Leftrightarrow P\ge1\)

\(\LeftrightarrowĐpcm.\)

Dấu " = " xảy ra khi \(a=b=c=1\)

Chúc bạn học tốt

có a³ kìa sao ko thay vào thành a^a+b+c r` giải thử nhỉ :D

Có: a²+b²+c²[tex]\geq[/tex]\(\dfrac{\left(a+b+c\right)^2}{3}\)

=>a²+b²+c²[tex]\geq[/tex]3;abc[tex]\leq[/tex]1(cô si 3 số)

[tex]P=\frac{a^{4}}{ab(2c+a)}+\frac{b^{4}}{bc(2a+b)}+\frac{c^{4}}{ac(2b+c)}[/tex]

=>P[tex]\geq[/tex][tex]\frac{(a^{2}+b^{2}+c^{2})^{2}}{6abc+abc(a+b+c)}[/tex]

P[tex]\geq[/tex][tex]\frac{3^{2}}{9abc}[/tex]

=[tex]\frac{1}{abc}[/tex]=1