{"id":5322,"date":"2025-10-08T23:00:00","date_gmt":"2025-10-09T03:00:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=5322"},"modified":"2025-10-09T01:14:12","modified_gmt":"2025-10-09T05:14:12","slug":"algebra-challenge-2025-10-09","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=5322","title":{"rendered":"Algebra Challenge – 2025\/10\/09"},"content":{"rendered":"\n

For number pairs $(r_i, c_i)$, $i=1,2,\u2026,n$, where $n\\ge 1$, $r_i\\ge 0$ and $c_i\\ge 0$, they have the following property:
$$\\sum_{i=1}^{n}r_i=\\sum_{i=1}^{n}c_i=n^2$$ Additionally, there exists a positive value of $k$ so that for every $i$ value,
$i=1,2,\u2026,n$, the following three inequalities hold:
$$r_i\\le \\dfrac{n^3}{k}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ c_i\\le \\dfrac{n^3}{k}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ k\\cdot(r_i+c_i)-r_i\\cdot c_i\\le n^3$$ Find the maximum value of $k$ in terms of $n$.\ud83d\udd11<\/a><\/p>\n

<\/p>\n\n\n\n

Solution:<\/strong> Assume that $k>\\dfrac{n(n+1)}{2}$, we have $$(k^2-n^3)-(k-n)^2=k^2-n^3-(n^2-2kn+k^2)=2nk-n^3-n^2$$ $$\\ \\ \\ \\ \\ \\ \\ >2n\\cdot\\dfrac{n(n+1)}{2}-n^3-n^2=0$$Therefore $$k^2-n^3>(k-n)^2\\ge 0$$ Rewriting $$k\\cdot(r_i+c_i)-r_i\\cdot c_i\\le n^3$$ as $$k^2-(k-r_i)(k-c_i)\\le n^3$$ i.e. $$(k-r_i)(k-c_i)\\ge k^2-n^3$$ We have $$(k-r_i)(k-c_i)>(k-n)^2\\ge 0\\tag{1}$$ Because $r_i\\le \\dfrac{n^3}{k}$, we have $$k-r_i\\ge k-\\dfrac{n^3}{k}=\\dfrac{k^2-n^3}{k}>0$$ i.e. $$k-r_i>0$$ Similarly, we have $$k-c_i>0$$ Because $n\\ge 1$, we have $$k-n>\\dfrac{n(n+1)}{2}-n=\\dfrac{n(n-1)}{2}\\ge 0$$ Based on $(1)$, we have $$\\sqrt{k-r_i}\\sqrt{k-c_i}>k-n$$ Therefore $$\\sum_{i=1}^{n}\\sqrt{k-r_i}\\sqrt{k-c_i}>n(k-n)\\tag{2}$$ By Cauchy-Schwarz inequality, we have $$\\Big{(}\\sum_{i=1}^{n}\\sqrt{k-r_i}\\sqrt{k-c_i}\\Big)^2\\le\\Big{(}\\sum_{i}^{n}(k-r_i)\\Big{)}\\cdot\\Big(\\sum_{i}^{n}(k-c_i)\\Big{)}\\ \\ \\ \\ \\ \\ \\ \\ \\ $$ $$=\\Big{(}nk-\\sum_{i=1}^{n}r_i\\Big{)}\\cdot\\Big{(}nk-\\sum_{i=1}^{n}c_i\\Big{)}=(nk-n^2)\\cdot(nk-n^2)=n^2(k-n)^2$$ Because $k-n>0$ and $n\\ge 1$, we have $$\\sum_{i=1}^{n}\\sqrt{k-r_i}\\sqrt{k-c_i}\\le n(k-n)$$ However, the above result contradicts $(2)$. Therefore the assumption of $k>\\dfrac{n(n+1)}{2}$ is false. We have
$$k\\le\\dfrac{n(n+1)}{2}$$ with the equality be attainable if we choose $r_i=c_i=n$, and $$k\\cdot(r_i+c_i)-r_i\\cdot c_i=n^3$$ for every value of $i$, $i=1,2,\u2026,n$. Therefore the maximum value of $k$ is $\\dfrac{n(n+1)}{2}$.<\/p>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

For number pairs $(r_i, c_i)$, $i=1,2,\u2026,n$, where $n\\ge 1$, $r_i\\ge 0$ and $c_i\\ge 0$, they have the following property:$$\\sum_{i=1}^{n}r_i=\\sum_{i=1}^{n}c_i=n^2$$ Additionally, there exists a positive value of $k$ so that for every $i$ value,$i=1,2,\u2026,n$, the following three inequalities hold:$$r_i\\le \\dfrac{n^3}{k}\\ \\ … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5322"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5322"}],"version-history":[{"count":15,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5322\/revisions"}],"predecessor-version":[{"id":5367,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5322\/revisions\/5367"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5322"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5322"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5322"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}