Proving a Quick and Easy Inequality

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This problem is from Algebraic Inequalities – Old And New Methods by Vasile Cirtoaje
#ChallengingMathProblems #ProvingInequalities
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[mrphlip]

Each of a, b, c, d is at most 2.
The expression: (a²*a+b²*b+c²*c+d²*d) / (a²+b²+c²+d²) can be viewed as a weighted average, where the values being averaged are a, b, c, d and the weights are the squares.
As this is an average, it cannot be larger than the maximum of a, b, c, d which means it cannot be larger than 2.
But since a²+b²+c²+d²=4 we can multiply through by that, and get that a³+b³+c³+d³ can be at most 8.

[snejpu]

Hmm... I was expecting some advanced examlpes of AM, GM inequalities. But that was simple indeed. How about 4-variable calculus though? : )

[PunmasterSTP]

Inequality? More like "unparalleled quality"! Your videos are always amazing; thanks so much for making and sharing them.

[wongjiasheng]

But a can be -ve, so if a^2<= 4 doesn’t it mean -2 < a< 2?

[yahav]

I thoroughly enjoy your videos, keep it up!!

[rachidbenmeziane]

so clair and so simple, really I like it !

[manivannansundarapandian]

Using vector algebra, it can be proved as follows:
Consider two vectors X and Y such that

X = [a, b, c, d];

Y = [a^2, b^2, c^2, d^2];


Using Schwartz inequality:

|X. Y| <= length(X). length(Y) ;

|X. Y| = a^3 + b^3 + c^3 + d^3;

length(X) = sqrt(a^2+ b^2+ c^2 + d^2) = sqrt(4) = 2.

length(Y) = sqrt(a^4+ b^4 c^4 + d^4) < sqrt( (a^2+ b^2 c^2 + d^2)^2) = 4

so, length(Y) < 4.

so, a^3 + b^3 + c^3 + d^3 <= 8.

[samhan]

wow very clever method, thank you for sharing.

[math]

very well explained, thanks for sharing

[techysubham]

More on combination problems like last one

[kapsleeds]

watching your videos make me feel dumb 😂🤣😂 even though I took math up to my 3rd university year. or maybe I'm just rusty since I graduated 14 years ago and haven't open a math book ever since

[呂永志]

Does the equality hold only when one of a, b, c, d is 2, the others are 0?

[srijanbhowmick]

pretty good soln
how about the floor problem i gave ? any ideas ?