450. Nota matematica: "Asupra unei inegalitati remarcabile".
by Virgil Nicula, Nov 6, 2016, 12:38 PM
In articolul "O inegalitate intr-un triunghi" aparut in G.M.B.,2001 pag. 146 se propun si se demonstreaza de catre profesor univ. Laurentiu Panaitopol urmatoarele doua inegalitati 
Vom prezenta demonstratiile acestora fara a utiliza relatiile trigonometrice. Reamintim cateva identitati cunoscute
(notatiile sunt standard).
Demonstratie LP1. Inegalitatea devine:
![$a^2[s(s-a)+(s-b)(s-c)]-4as(s-a)\sqrt{(s-b)(s-c)}+8(s-a)^2(s-b)(s-c)\ge 0\ \stackrel{2}{\iff}$](//latex.artofproblemsolving.com/c/7/c/c7cfe0fa663adc39c8b55b6b09e1b756fcc17365.png)
![$4as(s-a)\sqrt{(s-b)(s-c)}+(s-b)(s-c)\left[4s(s-a)+(b+c-a)^2\right]\ge 0\iff$](//latex.artofproblemsolving.com/0/b/6/0b69c89655f4c0cdef41feacd0d40ea24228e60f.png)
![$s(s-a)\left[a^2-4a\sqrt{(s-b)(s-c)}+ 4(s-b)(s-c)\right]+$](//latex.artofproblemsolving.com/d/5/f/d5f459adbb15babf8aef19e02c3702f9b7844332.png)
![$s(s-a)\left[a-2\sqrt{(s-b)(s-c)}\right]^2+$](//latex.artofproblemsolving.com/c/8/7/c87a0a6d96da3ca1b3bc3d439ccdd22227fc0127.png)
what is true.
Observatie.
Din lantul de relatii 
![$a\left[\sqrt{s(s-a)}-h_a\right]+2rs=$](//latex.artofproblemsolving.com/2/f/e/2feaeed740f261c7193a6261e3907a321a0a23a5.png)
rezulta 
![$\sum\left[aR+2r(s-a)\right]\ge $](//latex.artofproblemsolving.com/8/a/e/8aec678784baacfdb9b84cca19d3628d5d198568.png)
Aceasta inegalitate se poate obtine si altfel pentru un triunghi neobtuzunghic:
and 
Demonstratie LP2. Inegalitatea devine
![$2am_aS\le a^2\left[s(s-a)+(s-b)(s-c)\right]+4(s-a)(s-b)(s-c)(s-2a)$](//latex.artofproblemsolving.com/8/0/1/801552f69a397dd751fdfa866cc9ce716f2c0042.png)
![$2am_aS\le a^2s(s-a)+(s-b)(s-c)\left[a^2+4(s-a)(s-2a)\right]\ \stackrel{2}{\iff}\ \boxed{a^2s(s-a)-2am_aS+(s-b)(s-c)(b+c-2a)^2\ge 0}\ (6).$](//latex.artofproblemsolving.com/8/8/d/88d36ed633df4406fc47cc29106456fadfb98c0e.png)
Cazul 1
Asadar, 
![$a^2s(s-a)-(s-b)(s-c)\left[2\left(b^2+c^2\right)-a^2\right]\ge 0\iff$](//latex.artofproblemsolving.com/3/0/c/30c380d59942c5b81e48ba167110c7df66304107.png)
![$a^2[s(s-a)+(s-b)(s-c)]-2(s-b)(s-c)\left(b^2+c^2\right)\ge 0\ | \odot 2\iff$](//latex.artofproblemsolving.com/b/a/8/ba8e78f5c858acaa9bc7a7a59e77544f173fd42b.png)
![$2a^2bc-\left(b^2+c^2\right)\left[a^2-(b-c)^2\right]\ge 0\iff$](//latex.artofproblemsolving.com/5/f/3/5f372afba978ba52ad1cab49ea6ba3f9966e4cbb.png)
what is true.
Observatie. Remarcam identitatea
![$4\left[bcm_a^2-2as(s-a)^2\right].$](//latex.artofproblemsolving.com/d/6/4/d644b74ced2b77cfd896fe386160c40bcfd865c9.png)
Intr-adevar, ![$E=(s-b)(s-c)\left[\left(b^2+c^2\right)-2bc\right]+$](//latex.artofproblemsolving.com/7/2/e/72e0c6008ba2a6b13f8d03ae45ee8adf614d7e41.png)
![$s(s-a)\left[\left(b^2+c^2\right)+2bc-4a(b+c)+4a^2\right]=$](//latex.artofproblemsolving.com/a/5/7/a572c78f9cbdf252f1c48709613ffdec3087bdac.png)
![$\left[s(s-a)+(s-b)(s-c)\right]\left(b^2+c^2\right)+$](//latex.artofproblemsolving.com/2/5/a/25a3e961fadb798d8b560ff1c256e09f163de15b.png)
![$2bc\left[(s-a)-(s-b)(s-c)\right]-$](//latex.artofproblemsolving.com/3/9/9/39999a229c6100f296731149bd2f7f64cd8f433f.png)
![$4as(s-a)\left[(b+c)-a\right]=$](//latex.artofproblemsolving.com/e/f/9/ef9ed683f026bc93f6e096da02a4eab4cdf9874f.png)
![$bc\left[2\left(b^2+c^2\right)-a^2\right]-8as(s-a)^2=$](//latex.artofproblemsolving.com/9/f/8/9f80f5734a0cdb888052e0c1122bc5c903be9ca4.png)
Deoarece 
si 
adica 
rezulta inegalitatea 
cu egalitate
este echilateral. Relativ la mediane reamintim cateva inegalitati cunoscute 
(notatiile sunt standard).
Vom prezenta demonstratiile acestora fara a utiliza relatiile trigonometrice. Reamintim cateva identitati cunoscute 
(notatiile sunt standard).Demonstratie LP1. Inegalitatea devine:
![$a^2[s(s-a)+(s-b)(s-c)]-4as(s-a)\sqrt{(s-b)(s-c)}+8(s-a)^2(s-b)(s-c)\ge 0\ \stackrel{2}{\iff}$](http://latex.artofproblemsolving.com/c/7/c/c7cfe0fa663adc39c8b55b6b09e1b756fcc17365.png)
![$4as(s-a)\sqrt{(s-b)(s-c)}+(s-b)(s-c)\left[4s(s-a)+(b+c-a)^2\right]\ge 0\iff$](http://latex.artofproblemsolving.com/0/b/6/0b69c89655f4c0cdef41feacd0d40ea24228e60f.png)
![$s(s-a)\left[a^2-4a\sqrt{(s-b)(s-c)}+ 4(s-b)(s-c)\right]+$](http://latex.artofproblemsolving.com/d/5/f/d5f459adbb15babf8aef19e02c3702f9b7844332.png)
![$s(s-a)\left[a-2\sqrt{(s-b)(s-c)}\right]^2+$](http://latex.artofproblemsolving.com/c/8/7/c87a0a6d96da3ca1b3bc3d439ccdd22227fc0127.png)
what is true.Observatie.
Din lantul de relatii 
![$a\left[\sqrt{s(s-a)}-h_a\right]+2rs=$](http://latex.artofproblemsolving.com/2/f/e/2feaeed740f261c7193a6261e3907a321a0a23a5.png)
rezulta 
![$\sum\left[aR+2r(s-a)\right]\ge $](http://latex.artofproblemsolving.com/8/a/e/8aec678784baacfdb9b84cca19d3628d5d198568.png)
Aceasta inegalitate se poate obtine si altfel pentru un triunghi neobtuzunghic:
and 
Demonstratie LP2. Inegalitatea devine
![$2am_aS\le a^2\left[s(s-a)+(s-b)(s-c)\right]+4(s-a)(s-b)(s-c)(s-2a)$](http://latex.artofproblemsolving.com/8/0/1/801552f69a397dd751fdfa866cc9ce716f2c0042.png)
![$2am_aS\le a^2s(s-a)+(s-b)(s-c)\left[a^2+4(s-a)(s-2a)\right]\ \stackrel{2}{\iff}\ \boxed{a^2s(s-a)-2am_aS+(s-b)(s-c)(b+c-2a)^2\ge 0}\ (6).$](http://latex.artofproblemsolving.com/8/8/d/88d36ed633df4406fc47cc29106456fadfb98c0e.png)
Cazul 1
Asadar, 
![$a^2s(s-a)-(s-b)(s-c)\left[2\left(b^2+c^2\right)-a^2\right]\ge 0\iff$](http://latex.artofproblemsolving.com/3/0/c/30c380d59942c5b81e48ba167110c7df66304107.png)
![$a^2[s(s-a)+(s-b)(s-c)]-2(s-b)(s-c)\left(b^2+c^2\right)\ge 0\ | \odot 2\iff$](http://latex.artofproblemsolving.com/b/a/8/ba8e78f5c858acaa9bc7a7a59e77544f173fd42b.png)
![$2a^2bc-\left(b^2+c^2\right)\left[a^2-(b-c)^2\right]\ge 0\iff$](http://latex.artofproblemsolving.com/5/f/3/5f372afba978ba52ad1cab49ea6ba3f9966e4cbb.png)
what is true.Observatie. Remarcam identitatea
![$4\left[bcm_a^2-2as(s-a)^2\right].$](http://latex.artofproblemsolving.com/d/6/4/d644b74ced2b77cfd896fe386160c40bcfd865c9.png)
Intr-adevar, ![$E=(s-b)(s-c)\left[\left(b^2+c^2\right)-2bc\right]+$](http://latex.artofproblemsolving.com/7/2/e/72e0c6008ba2a6b13f8d03ae45ee8adf614d7e41.png)
![$s(s-a)\left[\left(b^2+c^2\right)+2bc-4a(b+c)+4a^2\right]=$](http://latex.artofproblemsolving.com/a/5/7/a572c78f9cbdf252f1c48709613ffdec3087bdac.png)
![$\left[s(s-a)+(s-b)(s-c)\right]\left(b^2+c^2\right)+$](http://latex.artofproblemsolving.com/2/5/a/25a3e961fadb798d8b560ff1c256e09f163de15b.png)
![$2bc\left[(s-a)-(s-b)(s-c)\right]-$](http://latex.artofproblemsolving.com/3/9/9/39999a229c6100f296731149bd2f7f64cd8f433f.png)
![$4as(s-a)\left[(b+c)-a\right]=$](http://latex.artofproblemsolving.com/e/f/9/ef9ed683f026bc93f6e096da02a4eab4cdf9874f.png)
![$bc\left[2\left(b^2+c^2\right)-a^2\right]-8as(s-a)^2=$](http://latex.artofproblemsolving.com/9/f/8/9f80f5734a0cdb888052e0c1122bc5c903be9ca4.png)
Deoarece 
si 
adica 
rezulta inegalitatea 
cu egalitate
este echilateral. Relativ la mediane reamintim cateva inegalitati cunoscute 
(notatiile sunt standard).This post has been edited 77 times. Last edited by Virgil Nicula, Mar 13, 2019, 12:30 PM
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