Dễ CM : 

\(1< A< 2\)

mệt !

mik đăng lên bởi mik ko biết làm 

bn nói vậy mình ko hỉu 

làm giúp mik ik

mik đag cần bài này để ôn thi !

ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2018^2}< \frac{1}{2017.2018}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

                                                                         \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

                                                                           \(=1-\frac{1}{2018}=\frac{2017}{2018}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{2017}{2018}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\) không phải số tự nhiên

\(\Rightarrow1+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)\) là hỗn số

\(\Rightarrow A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\) không phải số tự nhiên ( đ p c m)

Lời giải:

$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$

$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$

Trừ theo vế:

\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)

Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$

$=4-\frac{6061}{4^{2019}}< 4$

$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)

\(\frac{a^4+b^4}{a^3+b^3}+\frac{b^4+c^4}{b^3+c^3}+\frac{c^4+a^4}{c^3+a^3}\ge2018\)

\(\Leftrightarrow\frac{a^4+b^4}{a^3+b^3}+\frac{b^4+c^4}{b^3+c^3}+\frac{c^4+a^4}{c^3+a^3}\ge a+b+c\)

\(\LeftrightarrowΣ_{cyc}\frac{a^3\left(a-c\right)+b^3\left(b-c\right)}{a^3+b^3}\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(a-b\right)\left(\frac{a^3}{c^3+a^3}-\frac{b^3}{b^3+c^3}\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\frac{c^3\left(a^2+ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)\left(b+c\right)\left(b^2-bc+c^2\right)}\right)\ge0\)

BĐT cuối cùng liếc qua cũng biết thừa đúng :) nên ta có ĐPCM

Dấu "=" <=> a=b=c 

Ủng hô va` kb với mình nhé ^^

Bài này làm dài lắm

A/B>1/2018

\(\frac{A}{B}>\frac{1}{2018}\)

id nhu 1 tro dua

\(\frac{3}{4}A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^4+...-\left(\frac{3}{4}\right)^{2018}+\left(\frac{3}{4}\right)^{2019}\)

\(\frac{3}{4}A+A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^4+...-\left(\frac{3}{4}\right)^{2018}+\left(\frac{3}{4}\right)^{2019}+1-\frac{3}{4}+\left(\frac{3}{4}\right)^2...\)( Bn tự ghi lại A do máy mình ko đủ độ rộng )

\(\frac{7}{4}A=\left(\frac{3}{4}\right)^{2019}+1\)

\(A=\text{ }\left[\left(\frac{3}{4}\right)^{2019}+1\right]:\frac{7}{4}\)

\(A=\text{ }\frac{\left[\left(\frac{3}{4}\right)^{2019}+1\right].4}{7}\)

=> A là phân số

=> A ko phải số nguyên

Lời giải:

Xét hiệu:

\(2(a^4+c^4)-(a^3+c^3)(a+c)=2(a^4+c^4)-(a^4+a^3c+ac^3+c^4)\)

\(=a^4+c^4-a^3c-ac^3=(a-c)(a^3-c^3)=(a-c)^2(a^2+ac+c^2)\geq 0\)

với mọi \(a,c>0\)

Do đó: \(2(a^4+c^4)\geq (a^3+c^3)(a+c)\Leftrightarrow \frac{a^4+c^4}{a^3+c^3}\geq \frac{a+b}{2}\)

Hoàn toàn tương tự ta có:
\(\left\{\begin{matrix} \frac{b^4+c^4}{b^3+c^3}\geq \frac{b+c}{2}\\ \frac{a^4+b^4}{a^3+b^3}\geq \frac{a+b}{2}\end{matrix}\right.\)

Cộng theo vế các BĐT thu được:

\(\frac{a^4+b^4}{a^3+b^3}+\frac{b^4+c^4}{b^3+c^3}+\frac{c^4+a^4}{c^3+a^3}\geq \frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}=a+b+c=2018\)

Ta có đpcm.

Dấu bằng xảy ra khi $a=b=c=\frac{2018}{3}$

\(\dfrac{a^4+b^4}{a^3+b^3}+\dfrac{b^4+c^4}{b^3+c^3}+\dfrac{c^4+a^4}{c^3+a^3}\ge2018\)

\(\Leftrightarrow\dfrac{a^4+b^4}{a^3+b^3}+\dfrac{b^4+c^4}{b^3+c^3}+\dfrac{c^4+a^4}{c^3+a^3}\ge a+b+c\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^3\left(a-c\right)+b^3\left(b-c\right)}{a^3+b^3}\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)\left(\dfrac{a^3}{c^3+a^3}-\dfrac{b^3}{b^3+c^3}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{c^3\left(a^2+ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)\left(b+c\right)\left(b^2-bc+c^2\right)}\right)\ge0\)

Dễ thấy BĐT cuối luôn đúng nên ta có ĐPCM

Dấu "=" <=> \(a=b=c=\dfrac{2018}{3}\)

 help me