inequality

\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\)

\(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\)

\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

\( \dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14

\(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}\)

\(=1-\dfrac{1}{n^2}< 1\)

Ta có: \(\dfrac{2n+1}{n^2\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

Do đó \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

       \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

       \(=1-\dfrac{1}{\left(n+1\right)^2}< 1\)

Vậy \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\forall n\in\)N*

i have a solution but it's ugly

If 

 and    then we get a value  

We'll prove that it's a minimal value. Thus, we need to prove that

WLOG 

. Hence:

Similar

And 

Let 

 and we have:

By Cauchy-Schwarz's ine we have: 

Thus, it's enough to prove that 

It's true by AM-GM 

i have a solution but it's ugly

If \(a=b=1\) and \(c=0\)   then we get a value  \(2+\frac{1}{\sqrt2}\)

We'll prove that it's a minimal value. Thus, we need to prove that

\(\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

WLOG \(c=\min\{a,b,c\}\). Hence:

\(\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0\)

Similar\(\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0\)

And \(\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0\)

Let \(\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)\)\(\Rightarrow xy=1\) and we have:

\(x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}\)

\(\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}\)

\(\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2\)

By Cauchy-Schwarz's ine we have: 

\(\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y\)

\(=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2\)

Thus, it's enough to prove that \(\sqrt{x^2+y^2}+\sqrt{2}\ge2\)

It's true by AM-GM \(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)

tôi để lại số quả táo là:

My try: WLOG a≥b≥c hence we have: am−n≥bm−n≥cm−n

⇒anbn+cn≥bncn+an≥cnan+bn

By Chebyshev's inequality we have: 

∑ambn+cn=∑am−n⋅anbn+cn

≥am−n+bm−n+cm−n3⋅∑anbn+cn

≥∑am−n+bm−n+cm−n2

We're done when a=b=c

P/s: This's discuss not A4 (Auto Ask Auto Answer)

My try: WLOG \(a\ge b\ge c\) hence we have: \(a^{m-n}\ge b^{m-n}\ge c^{m-n}\)

\(\Rightarrow\dfrac{a^n}{b^n+c^n}\ge\dfrac{b^n}{c^n+a^n}\ge\dfrac{c^n}{a^n+b^n}\)

By Chebyshev's inequality we have: 

\(\text{∑}\dfrac{a^m}{b^n+c^n}=\text{∑}a^{m-n}\cdot\dfrac{a^n}{b^n+c^n}\)

\(\ge\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{3}\cdot\text{∑}\dfrac{a^n}{b^n+c^n}\)

\(\ge\text{∑}\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{2}\)

We're done when \(a=b=c\)

P/s: This's discuss not A⁴ (Auto Ask Auto Answer)

Considering the right expression we have  \(\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}\)

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

Apply Cauchy Schwarz inequality we have the formula \(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{1}{a+b}\)

Apply this formula to the following expression we have

\(\dfrac{a}{a+b+a+c}\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)\(;\dfrac{b}{a+b+b+c}\le\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)

\(\dfrac{c}{a+c+c+a}\le\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{c+a}\right)\)

\(\Rightarrow\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+c+a}\le\dfrac{a+b}{4\left(a+b\right)}+\dfrac{b+c}{4\left(b+c\right)}+\dfrac{c+a}{4\left(c+a\right)}=\dfrac{3}{4}\)

\(\Rightarrow abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}\)

We now need to prove that \(\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}\)

\(\Leftrightarrow\dfrac{a^3b}{3a+b}-\dfrac{abc}{4}+\dfrac{b^3c}{3b+c}-\dfrac{abc}{4}+\dfrac{c^3a}{3c+a}-\dfrac{abc}{4}\ge0\)

\(\Leftrightarrow\dfrac{a^3bc}{c\left(3a+b\right)}-\dfrac{abc}{4}+\dfrac{ab^3c}{a\left(3b+c\right)}-\dfrac{abc}{4}+\dfrac{c^3ab}{b\left(3c+a\right)}-\dfrac{abc}{4}\ge0\)

\(\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{b^2}{a\left(3b+c\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{1}{4}\right]\ge0\)

\(\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{3}{4}\right]\ge0\)

\(\Leftrightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}\)

Apply Cauchy Schwarz inequality Engel form we have 

\(\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\)

According to the Cauchy's consequence, we have \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{4\left(ab+bc+ac\right)}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}\)

So now we have \(\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}\)

But \(abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}\)

\(\Rightarrow\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}\) ( things must be proven.)

Yes but it's not mine, i need new method :)

Let x1=a2a1;x2=a3a2;...;xn=a1an

We need prove a1a1+a2+a3+a2a2+a3+a4+...+anan+a1+a2>1

It's right because n>3

 and ai+ai+1+ai+2<a1+a2+...+an∀i

Yes but it's not mine, i need new method :)

Let \(x_1=\dfrac{a_2}{a_1};x_2=\dfrac{a_3}{a_2};...;x_n=\dfrac{a_1}{a_n}\)

We need prove \(\dfrac{a_1}{a_1+a_2+a_3}+\dfrac{a_2}{a_2+a_3+a_4}+...+\dfrac{a_n}{a_n+a_1+a_2}>1\)

It's right because \(n>3\) and \(a_i+a_{i+1}+a_{i+2}< a_1+a_2+...+a_n\forall i\)

Ace Legona : you can share a answer of problem.

P=0 vì a+b+c=abc

suy ra abc -abc=0

answer :0

P=0 vì a+b+c=abc

suy ra abc -abc=0

answer :0

wrong tag : maxima-minima

That too long!

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

⇒1n3<1(n−1)n(n+1)

1(n−1)n(n+1)=1n.1(n−1)(n+1)

=1n.(n+1)−(n−1)(n−1)(n+1).12=12.1n.(1n−1−1n+1)

=12.(1(n−1)n−1n(n+1))

Now we have :

E < 12.3.4+13.4.5+14.5.6+...+1(n−1)n(n+1)

=12(12.3−13.4)+12(13.4−14.5)+12(14.5−15.6)+...+12(1(n−1)n−1n(n+1))

=12(12.3−1n(n+1))=112−12n(n+1)<112

Hence,E<112

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n² - 1) = n³ - n < n³

\(\Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{n}.\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=\dfrac{1}{n}.\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}.\dfrac{1}{2}=\dfrac{1}{2}.\dfrac{1}{n}.\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

Now we have :

E < \(\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+\dfrac{1}{2}\left(\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)+\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}< \dfrac{1}{12}\)

Hence,\(E< \dfrac{1}{12}\)