Brilliant Inequality Contest - Season 2

Welcome all to Season 2 of the Brilliant Inequality Contest. Like the Brilliant Integration Contest, the aim of the Inequality Contest is to improve skills and techniques often used in Olympiad-style Inequality problems. But above all, the main reason is so we can have fun.

Anyone is allowed to participate, as long as they adhere to the following rules.

  1. I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

  2. A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

  3. Please make a substantial comment. Spam or unrelated comments will be deleted. (To help me out, try to delete your comments within an hour of when you posted them if they are not the solutions.)

  4. Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 48 hours. Then, you must post the solution and you have the right to post a new problem.

  5. If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

  6. The scope of the problems is Olympiad-style inequalities, but non-standard inequalities are also allowed, (due to increasing number of non-standard inequalities found in olympiads)

  7. You are not allowed to post problems requiring calculus in the solutions (use of differentiation to prove a curve is concave or convex is allowed).

  8. Lagrange Multipliers are not allowed to be used in a solution.

  9. Most Standard Inequalities are allowed to be used, including AM-GM, Muirhead, Power Mean and Weighted Power Mean, Cauchy-Schwarz, Holder, Rearrangement, Chebyshev, Schur, Jensen, Karamata, Reverse Rearrangement, Bernoulli's Inequality and Titu's lemma.

  10. Try to post the simplest solution possible. For example, if someone posted a solution using Holder, Titu's and Cauchy, when there is a solution using only AM-GM, the latter is preferred.

Format your proof as follows:

SOLUTION OF PROBLEM (insert problem number here)

[Post your solution here]


PROBLEM (insert problem number here)

[Post your problem here]

Remember to reshare this note so it goes to everyone out there. Also remember to keep the comments sorted by Newest, so you can see the current problem. And above all else, have fun!!!


PROBLEM 1

For real numbers a1,a2,a_1, a_2, \ldots, if an1+an+12ana_{n-1} + a_{n+1} \geq 2a_n for n2n\geq 2, then prove that

An1+An+12An for n2.A_{n-1} + A_{n+1} \geq 2A_n \quad \text{ for } n\geq 2.

where AnA_n is the average of a1,a2,,ana_1, a_2, \ldots, a_n.

Note by Sharky Kesa
2 years, 6 months ago

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Problem 4

Prove that for all non-negative real numbers x,y,zx,y,z,

6(x+yz)(x2+y2+z2)+27xyz10(x2+y2+z2)3/26(x+y-z)(x^2+y^2+z^2)+27xyz \le 10(x^2+y^2+z^2)^{3/2}

Brilliant Member - 2 years, 6 months ago

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Applying Cauchy-Schwarz inequality, we have:

(x2+y2+z2)(22+22+12)(2x+2y+z)2\quad (x^2+y^2+z^2)(2^2+2^2+1^2)\ge(2x+2y+z)^2

or 3(x2+y2+z2)1/22x+2y+z3(x^2+y^2+z^2)^{1/2}\ge 2x+2y+z.

Hence,

10(x2+y2+z2)3/26(x+yz)(x2+y2+z2)\quad10(x^2+y^2+z^2)^{3/2}-6(x+y-z)(x^2+y^2+z^2)

(x2+y2+z2)(103(2x+2y+z)6(x+yz))\ge(x^2+y^2+z^2)\left(\dfrac{10}{3}(2x+2y+z)-6(x+y-z)\right)

=13(x2+y2+z2)(2x+2y+28z)=\dfrac{1}{3}(x^2+y^2+z^2)(2x+2y+28z)

Appying AM-GM inequality, we have:

x2+y2+z2=4x24+4y24+z29x8y8z2489x^2+y^2+z^2=4\cdot\dfrac{x^2}{4}+4\cdot\dfrac{y^2}{4}+z^2\ge9\sqrt[9]{\dfrac{x^8y^8z^2}{4^8}}

2x+2y+28z=2x+2y+74z92x2y(4z)79=948xyz792x+2y+28z=2x+2y+7\cdot 4z\ge 9\sqrt[9]{2x\cdot2y\cdot(4z)^7}=9\sqrt[9]{4^8xyz^7}

This implies that (x2+y2+z2)(2x+2y+28z)81xyz(x^2+y^2+z^2)(2x+2y+28z)\ge 81xyz, we are done.

The equality holds if and only if x=y=2zx=y=2z.

Khang Nguyen Thanh - 2 years, 6 months ago

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Problem 6:

Let aa, bb and cc be positive reals such that abc=1abc=1. Prove that: 1+(a3+1)(b3+1)(c3+1)31a+1b+1c1+\sqrt[3]{(a^3+1)(b^3+1)(c^3+1)}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}

Khang Nguyen Thanh - 2 years, 6 months ago

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Solution:

1+(a3+1)(b3+1)(c3+1)3ab+bc+ca(a3+1)(b3+1)(c3+1)(ab+bc+ca1)3cyca3+3cyca2b2+6cyca2bc3syma2b+3cycab+3\begin{aligned} 1+\sqrt[3]{(a^3+1)(b^3+1)(c^3+1)} &\geq \dfrac{a}{b} + \dfrac {b}{c} + \dfrac {c}{a}\\ (a^3+1)(b^3+1)(c^3+1) &\geq (ab+bc+ca-1)^3\\ \displaystyle \sum_{\text{cyc}} a^3 + 3 \displaystyle \sum_{\text{cyc}} a^2 b^2 + 6 \displaystyle \sum_{\text{cyc}} a^2 bc &\geq 3 \displaystyle \sum_{\text{sym}} a^2 b + 3 \displaystyle \sum_{\text{cyc}} ab + 3 \end{aligned}

The rest is just homogenisation and Muirhead bashing, which I'll update later since I have to go.

Sharky Kesa - 2 years, 6 months ago

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Can you update your complete solution?

Khang Nguyen Thanh - 2 years, 6 months ago

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let us change first n-1 -> let's say z then az a_{z} + az+2 a_{z+2} >= 2 az+1 a _{z+1} , now An A_{n} = a1 a_{1} ...........+ az a_{z} if we repeatedly add the given condition in the question i.e. az a_{z} + az+2 a_{z+2} >= 2 az+1 a _{z+1} you may refer to these pics but the handwriting is very poor :P!

Brilliant Member - 2 years, 6 months ago

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You have quite a few typos, but the essence of the proof is there, so I think you have solved it. I would recommend just going over the proof and fixing up the small errors you made to make it better. Good job. :)

Sharky Kesa - 2 years, 6 months ago

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For anyone interested in my proof of this problem, the algebra bash is fairly straightforward (clear out denominators, subtract and add, but be careful not to make errors). We manipulate An1+An+12AnA_{n-1} + A_{n+1} \geq 2A_{n} to

a1+a2++an1n2+n22an+n(n1)2an+10a_{1} + a_{2} + \ldots + a_{n-1} - \dfrac {n^2+n-2}{2} a_n + \dfrac {n(n-1)}{2} a_{n+1} \geq 0

However, we can convert this into the telescoping sum

=k=2nk(k1)2(ak12ak+ak+1)0=\displaystyle \sum_{k=2}^n \dfrac{k(k-1)}{2}(a_{k-1} - 2a_{k} + a_{k+1}) \geq 0

The final expression is true by our original condition, so we are done.

Sharky Kesa - 2 years, 6 months ago

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yes, good one !

Brilliant Member - 2 years, 6 months ago

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2ndques \large 2^{nd} ques if a,b,c are the sides of a triangle provided with perimeter of this triangle i.e 10 Prove that: \large Prove\ that : a2+b2+c2+25abc<50 \large { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+\frac { 2 }{ 5 } abc\quad <\quad 50

Brilliant Member - 2 years, 6 months ago

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Solution to Problem 2

Let a=x+ya=x+y, b=y+zb=y+z, c=z+xc=z+x. Since aa, bb and cc are sides of a triangle, xx, yy and zz must be positive by incentre properties. Also, since a+b+c=10a+b+c=10, x+y+z=5x+y+z=5. Substituiting this in to the inequation we wish to prove, we have:

(x+y)2+(y+z)2+(z+x)2+25(x+y)(y+z)(z+x)<505x2+5y2+5z2+5xy+5yz+5zx+x2y+x2z+y2x+y2z+z2x+z2y+2xyz<125(x+y+z)(x2+y2+z2+xy+yz+zx)+x2y+x2z+y2x+y2z+z2x+z2y+2xyz<(x+y+z)3x3+y3+z3+3(x2y+x2z+y2x+y2z+z2x+z2y)+5xyz<x3+y3+z3+3(x2y+x2z+y2x+y2z+z2x+z2y)+6xyz0<xyz\begin{aligned} (x+y)^2 + (y+z)^2 + (z+x)^2 + \dfrac{2}{5} (x+y)(y+z)(z+x) &< 50\\ 5x^2 + 5y^2 + 5z^2 + 5xy + 5yz + 5zx + x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 2xyz &< 125\\ (x+y+z)(x^2 + y^2 + z^2 + xy + yz + zx) + x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 2xyz &< (x+y+z)^3\\ x^3 + y^3 + z^3 + 3(x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 5xyz &< x^3 + y^3 + z^3 + 3(x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 6xyz\\ 0 &< xyz\\ \end{aligned}

We have already said that xx, yy and zz are positive, so the final expression is obviously true. Therefore, the first expression must be true. Thus, proven.


For further details, what I have done is use the incircle substitution so I can get rid of the weird triangle condition. Since every triangle has an incircle, and tangents to a circle from a point are equal, we can use this substitution.

In the second half of my proof, I homogenised all the terms by substituting x+y+z=5x+y+z=5 so they were all degree 3, and they simplified down to the final expression.

Sharky Kesa - 2 years, 6 months ago

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Yeah Ravi substitution is best when dealing with triangle sides inequalities !

Harsh Shrivastava - 2 years, 6 months ago

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Problem 3

Let aa, bb and cc be positive reals such that abc1abc \leq 1. Prove that

ac+ba+cba+b+c\dfrac {a}{c} + \dfrac {b}{a} + \dfrac {c}{b} \geq a + b + c

Sharky Kesa - 2 years, 6 months ago

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By AM-GM inequality,

ab+ca+ca3c2ab33c33=3c\dfrac{a}{b}+\dfrac{c}{a}+\dfrac{c}{a} \ge 3 \sqrt[3]{\frac{c^2}{ab}} \ge 3 \sqrt[3]{c^3} = 3c

Similarly, bc+ab+ab3a2bc33a33=3a\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{a}{b} \ge 3 \sqrt[3]{\frac{a^2}{bc}} \ge 3 \sqrt[3]{a^3} = 3a

and ca+bc+bc3b2ca33b33=3b\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{b}{c} \ge 3 \sqrt[3]{\frac{b^2}{ca}} \ge 3 \sqrt[3]{b^3} = 3b

Adding all the equaions, 3(ab+bc+ca)3(a+b+c)3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \ge 3(a+b+c).

Hence, ab+bc+caa+b+c\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge a+b+c with equality iff a=b=c=1a=b=c=1.

Brilliant Member - 2 years, 6 months ago

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Problem 5:

Let a,b,ca,b,c be real numbers such that a+b+c=0a+b+c=0.

Prove that:

a2b2+b2c2+c2a2+6abc+30a^2b^2+b^2c^2+c^2a^2+6abc+3\ge0

Khang Nguyen Thanh - 2 years, 6 months ago

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48 hours have passed since I posted this problem. Here is a solution.

WLOG, assume that ab0ab\ge 0.

From a+b+c=0a+b+c=0, we have c=(a+b)c=-(a+b).

Substituiting this in to the inequation we wish to prove, we have:

(a2+b2)(a+b)2+a2b26ab(a+b)+30\quad (a^2+b^2)(a+b)^2+a^2b^2-6ab(a+b)+3\ge0

or (a2+ab+b2)26ab(a+b)+30(a^2+ab+b^2)^2-6ab(a+b)+3\ge 0.

Appying AM-GM inequality, we have:

(a2+ab+b2)26ab(a+b)+3\quad (a^2+ab+b^2)^2-6ab(a+b)+3

(a2+ab+b2)23+2(a2+ab+b2)233ab[(a+b)2+4]2+3\ge \dfrac{(a^2+ab+b^2)^2}{3}+\dfrac{2(a^2+ab+b^2)^2}{3}-\dfrac{3ab[(a+b)^2+4]}{2}+3

9a2b23+2ab(a2+ab+b2)3ab(a+b)226ab+3\ge \dfrac{9a^2b^2}{3}+2ab(a^2+ab+b^2)-\dfrac{3ab(a+b)^2}{2}-6ab+3

=3(ab1)2+ab(ab)220=3(ab-1)^2+\dfrac{ab(a-b)^2}{2}\ge0.

We have equality for (a,b,c)=(1,1,2)(a,b,c)=(1,1,-2) and its permutations.

Khang Nguyen Thanh - 2 years, 6 months ago

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Problem 7:

Given distinct positive integers a1,a2,,ana_1, a_2, \ldots, a_n, prove that

(a17+a27++an7)+(a15+a25++an5)2(a13+a23++an3)2(a_1^7 + a_2^7 + \ldots + a_n^7) + (a_1^5 + a_2^5 + \ldots + a_n^5) \geq 2(a_1^3 + a_2^3 + \ldots + a_n^3)^2

Also, find the equality case.

Sharky Kesa - 2 years, 6 months ago

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Solution:

We try to prove this via induction.

For n=1n=1, we have

a17+a152(a13)2=a15(a11)20a_1^7 + a_1^5 - 2(a_1^3)^2 = a_1^5 (a_1-1)^2 \geq 0

so the base case n=1n=1 is true. Suppose the case is true for n=kn=k. For the case n=k+1n=k+1, we first WLOG a1<a2<<ak+1a_1 < a_2 < \ldots < a_{k+1}. We have

2(a13+a23++ak+13)22(a13+a23++ak3)2=2ak+16+4ak+13(a13++ak3)2ak+16+4ak+13(13+23++(ak+11)3)=2ak+16+4ak+13(ak+11)2ak+124=ak+17+ak+15\begin{aligned} 2(a_1^3+a_2^3+\ldots + a_{k+1}^3)^2 - 2(a_1^3+a_2^3+\ldots + a_k^3)^2 &= 2a_{k+1}^6 + 4a_{k+1}^3 (a_1^3 + \ldots + a_k^3)\\ & \leq 2a_{k+1}^6 + 4a_{k+1}^3 (1^3 + 2^3 + \ldots + (a_{k+1}-1)^3)\\ &= 2a_{k+1}^6 + 4a_{k+1}^3 \dfrac{(a_{k+1}-1)^2 a_{k+1}^2}{4}\\ &= a_{k+1}^7 + a_{k+1}^5\\ \end{aligned}

Thus, (a17+a27++an7)+(a15+a25++an5)2(a13+a23++an3)2(a_1^7 + a_2^7 + \ldots + a_n^7) + (a_1^5 + a_2^5 + \ldots + a_n^5) \geq 2(a_1^3 + a_2^3 + \ldots + a_n^3)^2, with equality iff a1=1,a2=2,,an=na_1 =1, a_2 = 2, \ldots, a_n = n.

Sharky Kesa - 2 years, 6 months ago

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Problem 8:

Let a1a2anan+1=0a_1 \geq a_2 \geq \ldots \geq a_n \geq a_{n+1} = 0 be a sequence of real numbers. Prove that

k=1nakk=1nk(akak+1)\sqrt {\displaystyle \sum_{k=1}^n a_{k}} \leq \displaystyle \sum_{k=1}^n \sqrt{k} (\sqrt{a_k} - \sqrt{a_{k+1}})

Sharky Kesa - 2 years, 6 months ago

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Solution:

Sorry for the late reply.

Let xk=akak+1x_k = \sqrt{a_k} - \sqrt{a_{k+1}}. Then, ak=(xk+xk+1++xn)2a_k = (x_k + x_{k+1} + \ldots + x_{n})^2. Thus,

k=1n=k=1n(xk+xk+1++xn)2=k=1nkxk2+21ijnixixjk=1nkxk2+21ijnijxixj=(k=1nkxk)2\begin{aligned} \displaystyle \sum_{k=1}^n &= \displaystyle \sum_{k=1}^n (x_k + x_{k+1} + \ldots + x_n)^2 = \displaystyle \sum_{k=1}^n kx_k^2 + 2 \displaystyle \sum_{1 \leq i \leq j \leq n} i x_i x_j\\ &\leq \displaystyle \sum_{k=1}^n kx_k^2 + 2\displaystyle \sum_{1\leq i \leq j \leq n} \sqrt{ij}x_i x_j = \left ( \displaystyle \sum_{k=1}^n \sqrt{k} x_k \right ) ^2\\ \end{aligned}

Taking the square root of both sides gives us the desired inequality.

Sharky Kesa - 2 years, 6 months ago

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Problem 9:

For 0abcde0 \leq a \leq b \leq c \leq d \leq e and a+b+c+d+e=1a+b+c+d+e=1, show that

ad+dc+cb+be+ea15ad + dc + cb + be +ea \leq \dfrac {1}{5}

Sharky Kesa - 2 years, 6 months ago

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Solution:

Since, abcdea\le b\le c\le d\le e, we get a+ba+cb+dc+ed+ea+b\le a+c\le b+d\le c+e\le d+e.

Applying the Chebyshev's inequality, we have:

2(a+b+c+d+e)2\quad2(a+b+c+d+e)^2

=(a+b+c+d+e)((d+e)+(c+e)+(b+d)+(a+c)+(a+b))=\left(a+b+c+d+e\right)\left((d+e)+(c+e)+(b+d)+(a+c)+(a+b)\right)

5(ad+ae+bc+be+cb+cd+da+dc+ea+eb)\ge 5(ad+ae+bc+be+cb+cd+da+dc+ea+eb)

=10(ad+dc+cb+be+ea)=10(ad+dc+cb+be+ea)

Hence, ad+dc+cb+be+ea15(a+b+c+d+e)2=15ad+dc+cb+be+ea\le\dfrac{1}{5}(a+b+c+d+e)^2=\dfrac{1}{5}.

The equality holds when a=b=c=d=e=15a=b=c=d=e=\dfrac{1}{5}.

Khang Nguyen Thanh - 2 years, 6 months ago

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Problem 10:

Let a,b,c,d,ea,b,c,d,e be non-negative real numbers such that a+b+c+d+e=5a+b+c+d+e=5. Prove that: abc+bcd+cde+dea+eab5abc+bcd+cde+dea+eab\le5

Khang Nguyen Thanh - 2 years, 6 months ago

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Solution:

Assume e=min{a,b,c,d,e}e=\min\{a, b, c, d,e\}. Then AM-GM gives :

e(c+a)(b+d)+bc(a+de)e(5e)24+(52e)2275e(c + a)(b + d) + bc(a + d - e) \le \dfrac{e(5 - e)^2}{4}+\dfrac{(5 - 2e)^2}{27}\le 5

the last one being equivalent with: (e1)2(e+8)0(e - 1)^2(e + 8)\ge 0, which is true.

Khang Nguyen Thanh - 2 years, 5 months ago

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Problem 11:

Let a,ba,b and cc be positive reals such that a+b+c=1a+b+c=1. Prove that: a+(bc)24+b+c3\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt b+\sqrt c\le\sqrt3

Khang Nguyen Thanh - 2 years, 5 months ago

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Can you please write your solution? The time has passed.

Sharky Kesa - 2 years, 5 months ago

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Applying the Cauchy-Schwarz inequality, we have:

(a+(bc)24+b+c)2\quad\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt b+\sqrt c\right)^2

3[a+(bc)24+(b+c)24+(b+c)24]\le 3\left[a+\dfrac{(b-c)^2}{4}+\dfrac{\left(\sqrt b+\sqrt c\right)^2}{4}+\dfrac{\left(\sqrt b+\sqrt c\right)^2}{4}\right]

=3[a+(bc)2+2(b+c)+4bc4]=3\left[a+\dfrac{(b-c)^2+2(b+c)+4\sqrt{bc}}{4}\right]

=3[a+2(b+c)+(b+c)2(bc)2+4bc4]=3\left[a+\dfrac{2(b+c)+\left(\sqrt b+\sqrt c\right)^2\left(\sqrt b-\sqrt c\right)^2+4\sqrt{bc}}{4}\right]

3[a+2(b+c)+2(b+c)(bc)2+4bc4]\le 3\left[a+\dfrac{2(b+c)+2(b+c)\left(\sqrt b-\sqrt c\right)^2+4\sqrt{bc}}{4}\right]

3[a+2(b+c)+2(bc)2+4bc4]\le 3\left[a+\dfrac{2(b+c)+2\left(\sqrt b-\sqrt c\right)^2+4\sqrt{bc}}{4}\right]

3[a+2(b+c)+2(b+c)4]=3\le 3\left[a+\dfrac{2(b+c)+2(b+c)}{4}\right]=3, Q.E.D.

The equality occurs iff a=b=c=13a=b=c=\dfrac{1}{3}.

Khang Nguyen Thanh - 2 years, 5 months ago

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@Khang Nguyen Thanh Post the next problem as well.

Sharky Kesa - 2 years, 5 months ago

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Problem 12:

Let a,ba,b and cc be real numbers such that: 1a,b,c1-1\le a,b,c\le 1 and a+b+c=0a+b+c=0.

Prove that: a2+b4+c62a^2+b^4+c^6\le 2.

Khang Nguyen Thanh - 2 years, 5 months ago

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Can you post your solution?

I'm not sure whether this hould keep running, due to lack of participants. Thoughts?

Sharky Kesa - 2 years, 5 months ago

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Sorry for late reply. In a recent days, my family and I are welcoming the Vietnamese New Year.

This is the solution:

Because a2+b4+c6a2+b2+c2a^2+b^4+c^6\le a^2+b^2+c^2, we can prove a stronger inequality: a2+b2+c22a^2+b^2+c^2\le 2.

WLOG, assume that abca\le b\le c. hence 1a0c1-1\le a\le 0\le c\le 1.

We have: a(a+1)0a(a+1)\le 0 or a2aa^2\le -a; c(c1)0c(c-1)\le 0 or c2cc^2\le c.

If 1b0-1\le b\le 0, we get b2bb^2\le -b, so a2+b2+c2ab+c=2c2a^2+b^2+c^2\le -a-b+c=2c\le 2.

If 0b10\le b\le 1, we get b2bb^2\le b, so a2+b2+c2a+b+c=2a2a^2+b^2+c^2\le -a+b+c=-2a\le 2.

Equility holds when (a,b,c)=(1,0,1)(a,b,c)=(-1,0,1) and its permutations.

Khang Nguyen Thanh - 2 years, 5 months ago

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Sharky, is the rules number 9 are all the inequality formula that have been discovered ?

I'm just here to try problem and study XD

Even though i didn't solve one of them

Jason Chrysoprase - 2 years, 5 months ago

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@Jason Chrysoprase There are so many more inequalities. Those listed are just the standard dozen and some extra.

Sharky Kesa - 2 years, 5 months ago

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