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Fie \( m>n, m, n\in\mathbb{N}^* \) si \( f:\mathbb{R}\rightarrow\mathbb{R},f(x)=n2^{mx}-m2^{nx} \).
a) Demonstrati ca \( f(x)\geq f(0),\forall x \).
b) \( f \) este strict descrescatoare pe \( (-\infty,0] \) si strict crescatoare pe \( [0,\infty) \).

OLM Prahova 2009

Am sa postez solutia mea care mi se pare foarte "trasa de par" si sper sa vina altcineva cu ceva mai frumos.
a) Fixez \( x\in \mathbb{R} \) si notez \( 2^x=a \):
\( f(x)\geq f(0)\Leftrightarrow \frac{a^m-1}{m}\geq\frac{a^n-1}{n} \).
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*, g(k)=\frac{a^k-1}{k} \) si demonstrez ca e crescatoare.
\( \frac{a^k-1}{k}\leq\frac{a^{k+1}-1}{k+1}\Leftrightarrow 1+ka^{k+1}\geq(k+1)a^k \), dar din inegalitatea mediilor stim ca:
\( \frac{1+a^{k+1}+\dots+a^{k+1}}{k+1}\geq\sqrt[k+1]{a^{k(k+1)}}=a^k. \)
Deci \( g(m)\geq g(n)\Rightarrow f(x)\geq f(0) \).

b) Fixez \( x>y, x,y\in\mathbb{R} \) si notez \( 2^x=a,2^y=b \).
i) \( a>b\geq1 \)
\( f(x)>f(y)\Leftrightarrow \frac{a^m-b^m}{m}\geq\frac{a^n-b^n}{n} \).
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*,g(k)=\frac{a^k-b^k}{k} \) si demonstrez ca g e crescatoare.
\( g(k)<g(k+1)\Leftrightarrow ak\frac{a^k}{b^k}+k+1>(k+1)\frac{a^k}{b^k}+kb \) si notand \( \frac{a^k}{b^k}=t>1 \):
\( k(at+1-t-b)> t-1 \)
avem \( k\geq1 \) si
\( at+1-t-b\geq t-1\Leftrightarrow t(a-2)+2>b \) care este adevarata caci \( t(a-2)+2>a-2+2=a >b \).
Deci \( g(n)<g(m)\Rightarrow f(x)>f(y) \) deci \( f \) e strict crescatoare pe \( (0,\infty) \).

ii) \( 0<b<a\leq1 \)
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*,g(k)=\frac{a^k-b^k}{k} \) si demonstrez ca e descrescatoare.
\( g(k+1)<g(k)\Leftrightarrow k(\frac{a}{b})^ka+k+1<(k+1)(\frac{a}{b})^k+kb\Leftrightarrow(\frac{a}{b})^k>\frac{k+1-kb}{k+1-ka} \)
\( (\frac{a}{b})^k=(\frac{a}{b}-1+1)^k>k(\frac{a}{b}-1)+1>\frac{k+1-kb}{k+1-ka} \)
Ultima inegalitate e echivalenta cu \( k(\frac{a-b}{b})>\frac{k(a-b)}{k+1-ka}\Leftrightarrow k+1>ka+b \) care e adevarata caci \( 1>a>b \).
\( g(m)<g(n)\Rightarrow f(x)<f(y) \Rightarrow f \) descrescatoare pe \( (-\infty,0] \).

Am ramas surprins de cat de mult m-a chinuit problema asta cu toate ca e de locala si tot cred ca are o rezolvare mai simpla.

\( \odot\ \ \ \{m,n\}\subset \mathbb{N}^* \) , \( m\ >\ n\ ,\ 0\ <\ a\ \ne\ 1 \) \( \Longrightarrow\ \) \( \frac {a^m-1}{m}\ \>\ \frac {a^n-1}{n}\ (*)\ . \)

Metoda 1. Se arata usor ca \( (*)\ \Longleftrightarrow\ f(a)\equiv na^m-ma^n+m-n\ >\ 0\ . \) Vom aplica schema lui Horner pentru radacina \( a:=1 \) (dubla !).

\( \overline{\underline{\left\|\ \begin{array}{cccccccccccccc}
* & \ \ \ \ a^m\ \ \ \ & \ \ \ \ a^{m-1}\ \ \ \ & \ \ \ a^{m-2}\ \ \ \ & \ldots & \ \ \ \ a^{n+1}\ \ \ \ & \ \ \ \ a^n\ \ \ \ & \ \ \ \ a^{n-1}\ \ \ \ & \ \ \ \ a^{n-2}\ \ \ \ & \ldots & \ \ \ \ a^2\ \ \ \ & \ \ \ \ a^1\ \ \ \ & \ \ \ \ a^0\\\\\\\\\\
* & n & 0 & 0 & \ldots & 0 & -m & 0 & 0 & \ldots & 0 & 0 & m-n\\\\\\\\\\
= & === & === & === & === & === & === & === & === & === & === & === & ===\\\\\\\\\\
1 & n & n & n & \ldots & n & n-m & n-m & n-m & \ldots & n-m & n-m & \underline{\overline{\left|\ 0\ \right|}}\\\\\\\\\\
1 & n & 2n & 3n & \ldots & (m-n)n & (m-n)(n-1) & (m-n)(n-2) & (m-n)(n-3) & \ldots & m-n & \underline{\overline{\left|\ 0\ \right|}} & *\end{array}\ \ \right\|}} \)

In concluzie, \( f(a)=(a-1)^2\ \cdot\ \left[\ n\ \cdot\ \sum_{k=1}^{m-n}\ ka^{m-k-1}\ +\ (m-n)\ \cdot\ \sum_{k=0}^{n-2}\ (k+1)a^k\ \right]\ >\ 0\ . \)

Metoda 2. Se arata usor ca \( (*)\ \Longleftrightarrow\ (a-1)E\ >\ 0 \) , unde \( E\equiv na^n\sum_{k=0}^{m-n-1}a^k-(m-n)\sum_{k=0}^{n-1}a^k\ . \)

Cazul 1. \( \overline{\underline{\left\|\ 0\ <\ a\ <\ 1\ \right\|}} \) Pentru \( k\in\mathbb N^* \) avem \( a^k\ <\ 1 \) si \( E\ <\ na^n\sum_{k=0}^{m-n-1}1-(m-n)\sum_{k=0}^{n-1}a^k= \)

\( n(m-n)a^n-(m-n)\sum_{k=0}^{n-1}a^k=(m-n)\left(na^n-\sum_{k=0}^{n-1}a^k\right)=(m-n)\left(\sum_{k=0}^{n-1}a^n-\sum_{k=0}^{n-1}a^k\right)= \)

\( (m-n)\sum_{k=0}^{n-1}a^k\left(a^{n-k}-1\right)\ <\ 0\ \Longrightarrow\ E\ <\ 0\ \Longrightarrow\ (a-1)\cdot E\ >\ 0\ \Longleftrightarrow\ (*)\ . \)

Cazul 2. \( \overline{\underline{\left\|\ a\ >\ 1\ \right\|}} \) Pentru \( k\in\mathbb N^* \) avem \( a^k\ >\ 1 \) si \( E\ >\ na^n\sum_{k=0}^{m-n-1}1-(m-n)\sum_{k=0}^{n-1}a^k= \)

\( n(m-n)a^n-(m-n)\sum_{k=0}^{n-1}a^k=(m-n)\left(na^n-\sum_{k=0}^{n-1}a^k\right)=(m-n)\left(\sum_{k=0}^{n-1}a^n-\sum_{k=0}^{n-1}a^k\right)= \)

\( (m-n)\sum_{k=0}^{n-1}a^k\left(a^{n-k}-1\right)\ >\ 0\ \Longrightarrow\ E\ >\ 0\ \Longrightarrow\ (a-1)\cdot E\ >\ 0\ \Longleftrightarrow\ (*)\ . \)


\( \odot\ \ m\in\mathbb N^*\ \Longrightarrow\ \ \ \) \( \underline{\overline{\left\|\ \begin{array}{ccccc}
0<b<a<1 & \ \Longrightarrow\ & \frac {a^{m+1}-b^{m+1}}{m+1} & \ <\ & \frac {a^{m}-b^{m}}{m}\\\\\\\\\\
1<b<a & \ \Longrightarrow\ & \frac {a^{m+1}-b^{m+1}}{m+1} & \ >\ & \frac {a^{m}-b^{m}}{m}\end{array}\ \right\|}} \)

Notam \( E_m=m\left(a^{m+1}-b^{m+1}\right)-(m+1)\left(a^m-b^m\right)\ ,\ n\in\mathbb N^* \).

Se observa ca \( E_{m+1}-E_m=(m+1)\left[a^m(a-1)^2-b^m(b-1)^2\right]\ . \)

Cazul 1. \( 1\ <\ b\ <\ a\ \Longrightarrow\ \left|\ \begin{array}{ccc}
a^m & > & b^m\\\\\\\\\\
(a-1)^2 & > & (b-1)^2\end{array}\right|\ \Longrightarrow\ E_{m+1}\ >\ E_m\ (\forall )\ m\in\mathbb N^*\ \Longrightarrow \)

\( (\forall )\ m\in\mathbb N^* \) , \( E_m\ >\ E_1=(a-b)(a+b-2)\ >\ 0\ \Longrightarrow\ E_m\ >\ 0\ \Longrightarrow\ \frac {a^{m+1}-b^{m+1}}{m+1}\ >\ \frac {a^{m}-b^{m}}{m}\ . \)

Cazul 2. \( 0\ <\ b\ <\ a\ <\ 1\ . \) In acest caz nu mai putem proceda ca la cazul precedent.

Se observa ca \( m(1-a)+(1-b)>0 \) , adica \( m+1>ma+b\ \Longrightarrow\ \frac {a^{m+1}-b^{m+1}}{m+1}<\frac {a^{m+1}-b^{m+1}}{ma+b} \) .

Insa \( \ \frac {a^{m+1}-b^{m+1}}{ma+b}<\frac {a^m-b^m}{m}\ \Longleftrightarrow\ \left(a^m-b^m\right)-mb^{m-1}(a-b)\ >\ 0\ \Longleftrightarrow\ \)

\( a^{m-1}+a^{m-2}b+\ldots +ab^{m-2}+b^{m-1}-mb^{m-1}\ >\ 0\ \Longleftrightarrow \)

\( \left(a^{m-1}-b^{m-1}\right)+b\left(a^{m-2}-b^{m-2}\right)+\ldots +b^{m-2}\left(a-b\right)\ >\ 0 \) , ceea ce este adevarat.