How to prove a known inequality from a book - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:01:43Z http://mathoverflow.net/feeds/question/69226 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69226/how-to-prove-a-known-inequality-from-a-book How to prove a known inequality from a book Sunni 2011-07-01T03:13:42Z 2011-07-03T23:25:24Z <p>The following inequality is from page 125 of D.S. Mitrinovic, J. Pecaric, A.M. Fink, Classical and new inequalities in analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.</p> <p>If $a_i>0$, $b_i>0$ for $i=1,\cdots, n$ and $A=\frac{\max a_k}{\min a_k}$, $B=\frac{\max b_k}{\min b_k}$ with $\frac{1}{p}+\frac{1}{q}=1$, $p>1$. Then</p> <p>$$\left(\sum\limits_{i=1}^na_i^p\right)^{1/p}\left(\sum\limits_{i=1}^nb_i^q\right)^{1/q}\le \frac{1}{p^{1/p}}\frac{1}{q^{1/q}}\frac{A^pB^q-1}{(BA^p-A)^{1/q}(AB^q-B)^{1/p}}\sum\limits_{i=1}^na_ib_i$$</p> <p>My question is how to prove this inequality (The book does not contain a proof). Though this is a known result, I am expecting different proofs from interested readers. Hopefully this does not go far away from the principle of this forum.</p> <p><strong>Added</strong> There are already satisfactory answers below, but let me add one question </p> <p>If $a_{1i}>0$, $a_{2i}>0, \cdots, a_{ri}>0$ for $i=1,\cdots, n$ and $A_1=\frac{\max a_{1k}}{\min a_{1k}}$, $A_2=\frac{\max a_{2k}}{\min a_{2k}},\cdots, A_r=\frac{\max a_{rk}}{\min a_{rk}}$ with $\sum\limits_{i=1}^r\frac{1}{p_i}=1$, $p_i>1$. Then $$\left(\sum\limits_{i=1}^na_{1i}^{p_1}\right)^{1/p_1}\left(\sum\limits_{i=1}^na_{2i}^{p_2}\right)^{1/p_2}\cdots \left(\sum\limits_{i=1}^na_{ri}^{p_r}\right)^{1/p_r}\le M(A_1,\cdots, A_r)\sum\limits_{i=1}^na_{1i}a_{2i}\cdots a_{ri}$$</p> <p>What would $M(A_1,\cdots, A_r)$ be? </p> http://mathoverflow.net/questions/69226/how-to-prove-a-known-inequality-from-a-book/69297#69297 Answer by Fedor Petrov for How to prove a known inequality from a book Fedor Petrov 2011-07-01T21:46:52Z 2011-07-01T21:46:52Z <p>sorry, it is not full proof, but too long for comment</p> <p>By scaling argument, we may suppose $a_i\in [1,A]$, $b_i\in [1,B]$. Note that the difference LHS-RHS is convex function in each $a_i$ and $b_i$ (the function $f(x)=(x^p+M)^{1/p}$ for constants $M>0$ and $p>1$ is convex on $(0,\infty)$). Hence it maximum on a closed segment is attained on one of endpoints. So, without loss of generality $a_i\in \{1,A\}, b_i\in\{1,B\}$. Also, for fixed arrays of $a_i$'s and $b_i$'s the RHS is minimal if large $a$'s are coupled with small $b$'s. It reduces our problem to the 2-parametric problem: number of ones in arrays of $a_i$'s, $b_i$'s. I guess that in the optimal situation number of ones between $a_i$'s equals the number of $B$'s between $b_i$'s, but I do not see any clear proof.</p> http://mathoverflow.net/questions/69226/how-to-prove-a-known-inequality-from-a-book/69300#69300 Answer by Gjergji Zaimi for How to prove a known inequality from a book Gjergji Zaimi 2011-07-01T23:01:13Z 2011-07-01T23:50:34Z <p>Here's how one can finish off Fedor's solution. We have $1\le k,r \le n$ and $A,B \geq 1$. We have to prove $$\left(p(AB^q-B)(n-k+kA^p)\right)^{1/p}\left(q(BA^p-A)(n-r+rB^p)\right)^{1/q}\le (A^pB^q-1)S$$ where $S=Ak+Br+(n-k-r)$ when $n\geq k+r$, and $S=B(n-k)+A(n-r)+AB(k+r-n)$ when $k+r\geq n$. </p> <p>Using Young's inequality $x^{1/p}y^{1/q}\le \frac{x}{p}+\frac{y}{q}$ on LHS (and after simplifying the expression) we reduce to proving $$(A^p-A)(B^q-B)\geq (A-1)(B-1)$$ in the case $n\geq k+r$, and $$(A-1)(B-1)\geq (1-\frac{1}{A^{p-1}})(1-\frac{1}{B^{q-1}})$$ in the other case. </p> <p>The first one follows from Bernoulli's inequality, $A^p-A=(1+(A-1))^p-A\geq 1+p(A-1)-A=(p-1)(A-1)$, and similarly $B^q-B\geq (q-1)(B-1)$, now notice that $(p-1)(q-1)=1$.</p> <p>The second case follows from the arithmetic-geometric mean, we have that $(p-1)A+\frac{1}{A^{p-1}}\geq p$ so $1-\frac{1}{A^{p-1}}\le (p-1)(A-1)$, and similarly for $B$, now just take the product.</p> http://mathoverflow.net/questions/69226/how-to-prove-a-known-inequality-from-a-book/69302#69302 Answer by GH for How to prove a known inequality from a book GH 2011-07-02T00:45:51Z 2011-07-02T01:39:34Z <p>The following proof was inspired by Fedor Petrov's and Gjergji's Zaimi's argument, but it is simpler.</p> <p>By a scaling argument we may assume $a_i\in[1,A]$, $b_i\in[1,B]$. The inequality can be rewritten as $$x^{1/p}y^{1/q} \leq (A^pB^q-1)\sum_{i=1}^n a_ib_i,$$ where $$x:=p(AB^q-B)\sum_{i=1}^na_i^p\qquad\text{and}\qquad y:=q(BA^p-A)\sum_{i=1}^nb_i^q.$$ By Young's inequality $x^{1/p}y^{1/q}\le \frac{x}{p}+\frac{y}{q}$, the above follows from $$\frac{x}{p}+\frac{y}{q}\leq (A^pB^q-1)\sum_{i=1}^n a_ib_i.$$ Therefore it suffices to show, for any $i$, $$(AB^q-B)a_i^p+(BA^p-A)b_i^q\leq (A^pB^q-1)a_ib_i.$$ The difference LHS-RHS is a convex function of $a_i$ and $b_i$, hence we can assume that <code>$a_i\in\{1,A\}$</code>, <code>$b_i\in\{1,B\}$</code>. The inequality becomes an identity when exactly one of $a_i$ and $b_i$ equals 1, while in the other two cases it is equivalent to $$(1-A^{1-p})(1-B^{1-q})\leq(A-1)(B-1)\leq (A^p-A)(B^q-B).$$ By convexity again, $$1-A^{1-p}\leq(p-1)(A-1)\leq A^p-A,$$ $$1-B^{1-q}\leq(q-1)(B-1)\leq B^q-B,$$ whence the required inequality follows upon noting that $(p-1)(q-1)=1$.</p> http://mathoverflow.net/questions/69226/how-to-prove-a-known-inequality-from-a-book/69325#69325 Answer by Fedor Petrov for How to prove a known inequality from a book Fedor Petrov 2011-07-02T13:17:32Z 2011-07-02T13:56:46Z <p>It is less or more the same as GH's proof, but let me explain how may one naturally come to such argument even without a priori knowing the constant. I do not refer here to other comments.</p> <p>At first, by standard scaling argument, $a_i\in [1,A]$, $b_i\in [1,B]$. Let's try to estimate $\sum a_ib_i$ from below via $\sum a_i^p$ and $\sum b_i^q$. The easiest way is by summping up inequalities $a_ib_i\geq \alpha a_i^p+\beta b_i^q$ for some positive constants $\alpha$, $\beta$. This inequality may be rewritten as $1\geq \alpha x+\beta x^{-q/p}$, where $x=a_i^{p-1}/b_i$. Since the RHS is convex in $x$, it suffices to check for maximal and minimal possible values of $x$, which corresponds to minimal $a_i$ and maximal $b_i$ or viceversa. In other words, we need to check two inequalities $B\geq \alpha+\beta B^q$, $A\geq \alpha A^p+\beta$, which correspond to pairs $(a_i,b_i)=(1,B)$ and $(a_i,b_i)=(A,1)$. It is natural to take $\alpha$, $\beta$ so that both inequalities are equalities. This is $2\times 2$ system, we solve it to find $\alpha=(AB^q-B)/(B^qA^p-1)$, $\beta=(BA^p-A)/(B^qA^p-1)$. Now it remains to get $$\sum a_ib_i\geq \alpha\sum a_i^p+\beta\sum b_i^q\geq (\alpha p)^{1/p}(\beta q)^{1/q} (\sum a_i^p)^{1/p}(\sum b_i^q)^{1/q}.$$</p>