Andrew Tate SOLVES math Olympiads problem| Level 1

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[@uzairname]

Do a matrix problem. Heard he has a lot to say about those

[@nourelhoudasabbagh]

Thank you, this very helpful 👏🏻👏🏻

[@yugam6578]

Andrew tate, which books do you recommend for maths Olympiad

[@philosophieoverdose9332]

Again thank you soo much for this video , at the same time i follow some books dedicated to olympiad inequalities, Today i want to share this inequalitie with the beautiful community,if:a,b,c are positif numbers and: a+b+c=1 proof that : 1/(b+c) +1/(a+c) +1/(a+b)>=9/2

[@MathWisdom42]

Nice one, thank you. Spoiler alert , a hint down below: . . . . . . . . . . Think of Cauchy-Schwarz inequality

[@philosophieoverdose9332]

@@MathWisdom42 i prof it using Nesbitt inequality , i'l try to do it by Cauchy Schwartz, again thank you soo much ,😊😊🌹

[@mr.lego-ist8904]

Keep on releasing these kinds of stuff that uses AI voices of popular people ✨... As a math enthusiast and a supporter of masculinity, this is 'perfect' content 💯...

[@latins4931]

You could also multiply both by x and have: x^2 + y^2 >= 2xy Subtracting both sides by 2xy you get: x^2 - 2xy + y^2 >= 0 And finally this equals to (x-y)^2 >= 0 The left term is a square which is always not negative, so the equation is true.

[@myst3ry_g3n3ral]

was my idea as well. i dont understand why they needed that arithmetic-geometric-formula this is sooooo much easier xd

[@testx0170]

Last inequality is just a+b≥2sqrt ab, cyclic product gives (a+b)(a+c)(b+c)≥8sqrt(a^2b^2c^2)=8abc, equality only holds when a=b, a=c, b=c or just a=b=c