Topics of studying inequalities 22

After my first basic par here , in this note, I will try to emphasis on Cauchy Schwarz inequality. In my opinion, this inequality is one of the most widely used and the most elementry of all inequalities (here by elementry I do not mean easy, but it is root of many other inequalities). So here we start of with Cauchy Schwarz inequality.. First we will see a short form of this inequality. For any real a,b,c,da,b,c,d we have (a2+b2)(c2+d2)=(ac+bd)2+(bcad)2(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(bc-ad)^2 So we get the inequality that (a2+b2)(c2+d2)(ac+bd)2(a^2+b^2)(c^2+d^2)≥(ac+bd)^2 The more generalized format can be written as (a12+a22+...an2)(b12+b22+....bn2)(a1b1+a2b2+......anbn)2(a_1^2+a_2^2+...a_n^2)(b_1^2+b_2^2+....b_n^2)≥(a_1b_1+a_2b_2+......a_nb_n)^2 The proof can be found here

Now, what type of problems can be designed on this topic, or else how to tackle these type on inequality. This problem is faced by a lot of people. So, whenever you see a problem, first try to relate it with AM-GM Inequality. Last note, I gave some good problems on AM-GM. Then Try to use Cauchy Schwarz . So, first Idea that I want to give a hint, based on my observations on various questions, I would like to add a tip here(Maybe it wont work all times). If ever you see a problem involving fractions i.e. Having a,b,c,,,a,b,c,,, on both numerator and denominator, with the numerator a perfect square, then you can apply Cauchy Schwarz as such a2b+c2d+.....(a+c+.......)2b+d+....\frac{a^2}{b}+\frac{c^2}{d}+.....≥\frac{(a+c+.......)^2}{b+d+....} The above is a direct applicaton of Cauchy Schwarz (Can you prove it?)

So, how Do we solve some inequalities involving Cauchy Schwarz ? Lets See

EXAMPLE;1EXAMPLE;1 Show that for all positive x,y,zx,y,z, We have x+y+z2[x2y+z+y2x+z+z2x+y]x+y+z≤2[\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}]

SOLUTIONSOLUTION:: Remember the point I raised in my notes. So, Now considering the right side of the inequality

2[x2y+z+y2x+z+z2x+y]2((x+y+z)22(x+y+z)x+y+z2[\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}]≥2(\frac{(x+y+z)^2}{2(x+y+z)}≥x+y+z

So, we see that after a perfect application of Cauchy Schwarz Inequality, the problem remains Just 1 line proof question. So, lets see some more examples

EXAMPLE;2EXAMPLE;2. First I would like to mention that this problem is taken from a Iran MO. So ,let us see the power of inequality in this case

Given that x,y,z>1x,y,z>1 and 1x+1y+1z=2\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2, SO prove that

x+y+zx1+y1+z1\sqrt{x+y+z}≥\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}

Solution::Solution:: So, how would you approach this question. Well the to prove part of inequality seems too difficult with AM-GM. Also in the right side of the inequality we have x1x-1 type of terms. SO we clearly see that simple product wont help. SO lets take a glance towards the given part. 1x+1y+1z=2\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2.

So, we try to get a term of x1x-1 type from given. So ,we clearly see that x1x+y1y+z1z=1\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1

No take a look at the left part, there we have x+y+zx+y+z term. So see that from our above result if we try to apply x×x1x=x1x×\frac{x-1}{x}=x-1. SO now we get a clear idea to apply Cauchy Schwarz. Therefore, By Application Of Cauchy Schwarz Inequality, we get

x+y+z=(x+y+z)(x1x+y1y+z1z)(x1+y1+z1)2x+y+z=(x+y+z)(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})≥(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2

Which leads to

x+y+zx1+y1+z1\sqrt{x+y+z}≥\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}

Now, Here are some beautiful problems that you must try in order to skill this topic. Also, if possible, I will be posting of more inequalities, Maybe Holder's Inequality or Jensen's Inequality of Just Advanced Cauchy Schwarz Inequality.

Problem 11:: For a,b,c,x,y,za,b,c,x,y,z Positive reals, prove that (5ax+ay+bx+3by)2(5a2+2ab+3b2)(5x2+2xy+3y2)(5ax+ay+bx+3by)^2\leq (5a^2+2ab+3b^2)(5x^2+2xy+3y^2)

Problem 22:: (Generalized) Let a1,a2,.......ana_1,a_2,.......a_n be positive reals. Prove that

(1+a1)(1+a2)(1+an)(1+a1×a2×a3××ann)n ( 1 + a_1)(1+a_2) \ldots (1 + a_n) \geq ( 1 + \sqrt[n]{a_1 \times a_2 \times a_3 \times \ldots \times a_n})^n

Problem 33:: This is pretty good one. The solution( Of Mine) Looks very big and unsolvable, Bu dont stop bcz in end there yeilds easy simplification. This one is taken from KMO Winter Program Test . Prove that for a,b,c>0a,b,c>0

(a2b+b2a+c2a)(ab2+bc2+ca2)abc+(a3+abc)(b3+abc)(c3+abc)3\sqrt{(a^2b+b^2a+c^2a)(ab^2+bc^2+ca^2)}≥abc+\sqrt[3]{(a^3+abc)(b^3+abc)(c^3+abc)}

Apart from all these problems, you can also post your doubts or questions in the comments so as to proved more questions for more people too. Thanks :)

Note by Dinesh Chavan
5 years, 4 months ago

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@Calvin Lin , @Seyoung Kim , Can anyone please help me fix my latex error in last 2nd problem. It seems OK on other sites when I checked !!

Dinesh Chavan - 5 years, 4 months ago

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Dinesh I wanna learn inequalities wat shud I refer in addition to urs beautiful note

Kundan Patil - 4 years, 8 months ago

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Also you spelled "Jensen's" wrong. There are several errors that I would revise, such as mismatched parenthesis and random capitalization.

Finn Hulse - 5 years, 4 months ago

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Excellent note! Very helpful, to prove is the more beautiful thing to do on mathematics! Sharing!

Carlos David Nexans - 5 years, 2 months ago

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Thank You very much, I will now try to write further topics too !

Dinesh Chavan - 5 years, 2 months ago

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This is an excellent note, but you've spelled Schwarz wrong in 10 different places. There's no "t" before the z.

Finn Hulse - 5 years, 4 months ago

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HAHA. OK I Will make the Edit,

Dinesh Chavan - 5 years, 4 months ago

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