Proving This Elegant Inequality

[@andreben6224]

Oh Nesbitt's inequality. It's such a classic it has many proofs and can serve to introduce many techniques for proving inequalities :D AM-HM, AM-GM, Cauchy-Schwarz inequality and Titu's variation, rearrangement, Jensen and various variable substitutions.

[@randomjin9392]

Without AM-GM: set b+c = x, a+c = y, a+b = z, then a = (-x+y+z)/2, b = (x-y+z)/2, c = (x+y-z)/2 and the inequality turns into (-x+y+z)/x+(x-y+z)/y+(x+y-z)/z ≥ 3. Now we have 3 times the -1 on the LHS after separating the fractions, so after moving that to RHS and grouping terms we have (y/x+x/y)+(z/x+x/z)+(y/z+z/y) ≥ 6 which is true because for any positive number R we have R+1/R ≥ 2 since R²-2R+1 = (R-1)² ≥ 0.

[@chamsderreche5750]

Actually, you technically used the AM-GM because that very last step is where AM-GM comes from Ig it'd be more "AM-GM free" if you proven that r+1/r>2 using calculus

[@res5139]

This was soooo random!

[@randomjin9392]

@@chamsderreche5750 Well, no. AM-GM is based on what I'm using (i.e. that the square of a real number is non-negative) because during the proof of AM-GM we rely on that fact. So both AM-GM and the reciprocals sum are relying on the same fundamental which for me is sufficient to call it "without AM-GM".

[@gdtargetvn2418]

I have known only 5-6 methods to prove this Nesbitt's inequality, but I have never seen anything as creative as this. This should be the best method in my opinion. Also fun fact: There are more than 40 methods to prove Nesbitt inequality.

[@dwaraganathanrengasamy6169]

Can u show me those other methods as well... Very curious 😄

[@chamsderreche5750]

(I will be using > for greater than or equal) 1.Rearrangement inequality WLOG a>b>c Then a+b>a+c>b+c Therefore their reciprocals are oppositely ordered The series a>b>c and 1/(b+c)>1/(a+c)>1/(a+b) are in the same order in nesbitt's inequality so by rearrangement, this permutation gives a greater value or equal to any other permutation, therefore LHS>a/(a+b)+b/(b+c)+c/(c+a) And LHS>b/(a+b)+c/(b+c)+a/(c+a) By summing these 2 inequalities and knowing that (a+b)/(a+b)=1 2LHS>3 LHS>3/2 2.Chebyshev Taking the same process with the series in rearrangement and by chebyshev 3LHS>(a+b+c)(1/(a+b)+1/(a+c)+1/(b+c))=3+LHS Therefore LHS>3/2 3.Cauchy Shwarz (in 3 different ways) By adding 3 to LHS(1 to every fraction) and multiplying by 2 the inequality is equivalent to (a+b+a+c+b+c)(1/(a+b)+1/(a+c)+1/(b+c))>9 Which is true by cauchy shwarz 4.Multiplying the enumerator and denominator of a/(b+c) by a to get a^2/(ab+ac) and similarly for other fractions, then by titu's lemma a^2/(ab+ac)+b^2/(ab+bc)+c^2/(cb+ac)>(a+b+c)^2/2 (ab+ac+bc) Now we only need to prove that (a+b+c)^2>3 (ab+ac+bc) Which is equivalent to (a-b)^2+(a-c)^2+(b-c)^2>0 Which is obviously true 5.the inequality is homogeneous, set a+b+c=1, by adding 1 to every fraction and 3 to RHS we get (a+b+c)/(a+b)=1/(a+b) so the inequality is equivalent to 1/(a+b)+1/(b+c)+1/(c+a)>9/2 By titu's lemma LHS>(1+1+1)^2/2 (a+b+c)=9/2 6.by clearing the denominators the inequality is equivalent to 2 (a^3+b^3+c^3)>a^2b+b^2a+a^2c+c^2a+b^2c+c^2b bc that the numbers are positive (a-b)^2 (a+b)>0 Therefore a^3+b^3>a^2b+b^2a Similarly a^3+c^3>a^2c+c^2a b^3+c^3>b^2c+c^2b By summing the inequalities we get the desired result

[@bosorot]

@@dwaraganathanrengasamy6169 there is a wikipedia on this topic with all other methods , google up Nesbitt's inequality.

[@rajsingh8372]

Yeah me too pretty common for Olympiad students

[@keikaku9298]

Another way is to use AM-HM inequality. 3 / (1/(b+c) + 1/(a+c) + 1/(a+b))

[@digxx]

WLOG a>=b>=c. Set s=(a+b+c)/3, then the objective is (a-s)/(b+c)+(b-s)/(a+c)+(c-s)/(a+b)>=0. case 1, sa+c: Then clearly lhs>=(a-s)/2b+(b-s)/2b+(c-s)/2b>=0 case 2, s>b implies 2b=(a-s)/2b+(b-s)/2b+(c-s)/2b>=0.

[@ummwho8279]

Steele's "The Cauchy-Schwarz Masterclass" gives this as an exercise in exercise 5.6 (pg. 84). If you want, Engel's book "Problem Solving Strategies" gives 5 different methods of proof for Nesbitt's inequality (pgs. 162-168). Hope that helps!

[@吳沛洋-f8y]

elegant indeed

[@lubosdostal8523]

From simple inequality: x+ 1/x ≥ 2. Set x1=(a+b)/(a+c), x2=(a+c)/(b+c), and x3=(a+b)/(b+c). Simplify (x1 + 1/x1) + (x2 + 1/x2) + (x3 + 1/x3) ≥ 6 and you'll get the required inequality... [Btw. simplify (sqrt(x) - 1/sqrt(x))^2 ≥ 0 you'll prove the initial inequality]

[@fierydino9402]

Glad to learn this :)

[@res5139]

Unbelievably cool! You're a magician......!!

[@noahtaul]

You can show that a/(b+c)>=(8a-b-c)/(4(a+b+c)) because it’s equivalent to (2a-b-c)^2>=0. Adding the other inequalities up gives the inequality you asked for

[@advaykumar9726]

Add +1+1+1 both sides and then use am hm inequality

[@kemalkayraergin5655]

Its way too cleaver

[@pranavsawantcoder]

Nesbitt Inequality

[@f5673-t1h]

Note that this bound is achieved when a = b = c.

[@pranavsawantcoder]

Or in other words when each of the fractions = 1/2 Nice symmetry

[@jamesjames1549]

You can try using Sedrakyan's inequality: cyc(ai^2/bi)>=(cyc(ai))^2/(cyc(bi)) So then, cyc(a/(b+c)) = cyc(a^2/a(b+c)) >= (a+b+c)^2/cyc(a(b+c)) =(a+b+c)^2/(ab+bc+ca+ac+ba+cb) = (a+b+c)^2/(2(ab+bc+ac)). Since cyc(a^2+b^2-2ab)>=0, then cyc(a^2)>=cyc(ab) {also can use AM-GM or Muirhead inequality} Then (a+b+c)^2/(2(ab+bc+ac)) = (a^2+b^2+c^2+2ab+2bc+2ac)/(2(ab+bc+ac)) >= 3(ab+bc+ac)/2(ab+bc+ac)=3/2.

[@UneFenetreSurLeMonde]

One « semi-method » (you will understand soon why I called it semi-method) is too use symetry argument that is very obvious : the equation don't depend if we change any value for variables a, b and c by each other, so any reasonning that finds a value for « a » that will make the quantity of the first part of the inegality a maximum, we will be able to do the same thing for « b » or « c », then if a minimum exists, then it is for a=b=c, then a/2a+a/2a+a/2a=1/2+1/2+1/2=3/2 we check that's a minimum for example a=1, b=1, c=2, then we have 1/3+1/3+2/2=1+2/3=5/3(1,6)>3/2(1,5) but we just have shown that if a minimum exists then it is 3/2 but we didn't demonstrate that 3/2 is a minmum ;) strange isn't it ?

[@johnwquigley1811]

Good video! How did you write on the computer? What software/technology do you use? Thank you lots 🙏

[@chamsderreche5750]

Ig it's a graphic tablet, idk what program they used tho

[@jkid1134]

Pretty

[@Deathranger999]

OK, and for the next challenge, solve for when that expression equals 4. :)

[@mayoor357]

Well can we go by trigonometry here?

[@tridibeshsamantroy9837]

INMO question

[@pranavsawantcoder]

Probably even RMO, it's just an application of AM-GM-HM. This is now taught as a standard result for even PRMO kids

[@tridibeshsamantroy9837]

@@pranavsawantcoder yy actually it was tought as an class illustration by our maths teacher in coaching he told us the year and examination in which this question was asked that's why

[@pranavsawantcoder]

@@tridibeshsamantroy9837 Ah noice

[@CreativeMathProblems]

Let a+b+c = S and apply Jensen on x/(S-x)

[@bismir4147]

Nesbitt

[@benjaminkarazi968]

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