$6(\sum a^3) \ge \frac{3}{2}(\sum a^2)$ (Chebyshev) $\frac{1}{2}(\sum a^2) \ge \frac{1}{8}$ (Cauchy)

Easy. Murhed. Or $6(a^3+b^3+c^3+d^3)\ge(a^2+b^2+c^2+d^2)+\frac{1}{8}\Leftrightarrow$ $\sum_{cyc}(6a^3-a^2-\frac{1}{32}-\frac{5}{8}(a-\frac{1}{4}))\geq0\Leftrightarrow\sum_{cyc}(a-\frac{1}{4})^2(3a+1)\geq0.$

Write it as $48(a^3+b^3+c^3+d^3)\geq 8(a^2+b^2+c^2+d^2)(a+b+c+d)+(a+b+c+d)^3$ Then use Muirhead and Holder.

silouan wrote: Write it as $48(a^3+b^3+c^3+d^3)\geq 8(a^2+b^2+c^2+d^2)(a+b+c+d)+(a+b+c+d)^3$ Then use Muirhead and Holder. I think only Muirhead is already OK,isn't it? (Because I did it by Muirhead only,just expand all terms...)

With using both Muirhead and Holder you have not so much expand.That's all

What is muirhead

Beat wrote: What is Muirhead? Muirhead's theorem is a very famous generalization of AM-GM inequality.

Soarer wrote: $6(\sum a^3) \ge \frac{3}{2}(\sum a^2)$ (Chebyshev) Why?? I don't see Trebúsep here.

tkhtn wrote: Soarer wrote: $6(\sum a^3) \ge \frac{3}{2}(\sum a^2)$ (Chebyshev) Why?? I don't see Trebúsep here. By using Chebychev we have $(a+b+c+d)(a^2+b^2+c^2+d^2)\leq 4(a^3+b^3+c^3+d^3)$ .But $a+b+c+d=1$ etc

Oh my, sory . I don't remember: $a+b+c+d=1$

i've got a different solution from cauchy-schwarz we have that \[ \sum a^3=(\sum a)(\sum a^3 )\geq (\sum a^2)^2 \] then we just need to show that \[ 6x ^2-x-\frac{1}{8}\geq 0 \] where $x=\sum a^2$. but \[ 6x^2-x-\frac{1}{8}=\frac{1}{8}(12x+1)(4x-1) \] and from cauchy-schwarz again we have that $4x\geq (a+b+c+d)^2=1$

hy i wonder what were the results of the selection test? how many people solved 3 or more problems?

seems to be 1 full mark, several (should be ~5) 21 or above.

More interesting is the following: Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=4$. Prove that $5(a^2+b^2+c^2+d^2) \ge a^3+b^3+c^3+d^3+16$.

Hallo Vasile! How are you ? Our book is ready , but the last correction are still on the road.When will be ready your new book in Inequalities? Have a nice day - babis

Vasc wrote: More interesting is the following: Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=4$. Prove that $5(a^2+b^2+c^2+d^2) \ge a^3+b^3+c^3+d^3+16$. Write the inequality of the form $\sum (3-a)(a-1)^2 \ge 0$ If $a \leq 3$, done. Else, write the inequality as $(a+c+d)(b-1)^2 + (a+b+d)(c-1)^2 + (a+b+c)(d-1)^2 \ge (b-1)^2+(c-1)^2+(d-1)^2+(a-3)(a-1)^2$ By Cauchy-Schwarz, $(3a-9)((b-1)^2+(c-1)^2+(d-1)^2) \ge (a-3)(a-1)^2$. However, we also have $a - (3a-9) = 9-2a$, and $(9-2a)(b-1)^2 \ge (b-1)^2$ because $a \leq 4$, same for $c,d$. QED

First Solution: By Cauchy-Shwarz inequality, we have: (a^3+b^3+c^3+d^3)(a+b+c+d)=(a^3+b^3+c^3+d^3)>=(a^2+b^2+c^2+d^2)^2 (1). So,6(a^3+b^3+c^3+d^3)-(a^2+b^2+c^2+d^2+1/8)>=6(a^2+b^2+c^2+d^2)^2-(a^2+b^2+c^2+ +d^2+1/8)=(a^2+b^2+c^2+d^2-1/4)(6(a^2+b^2+c^2+d^2)+1/2)>=o,because sqrt((a^2+b^2+c^2+d^2)/4)>=(a+b+c+d)/4=1/4(as average square and AM), a^2+b^2+c^2+d^2 >=1/4; 6(6(a^2+b^2+c^2+d^2)+1/2)>=0. We are done. Second Solution: Look at function f(x)=6x^3-x^2-(5x-1)/8. f(x)=1/8(3x+1)(4x-1)^2. (1). For x=a,x=b,x=c and x=d, f(a),f(b),f(c),f(d)>=0, as 3a+1,3b+1,3c+1,3d+1>0, (4x-1)^2>=0. So, f(a)+f(b)+f(c)+f(d)>=0. Which is the same as the following: 6(a^3+b^3+c^3+d^3)-(a^2+b^2+c^2+d^2)-(5(a+b+c+d)-4)/8=6(a^3+b^3+c^3+d^3)-(a^2+b^2+c^2+d^2+1/8) >=0. We are done.

Vasc wrote: More interesting is the following: Let $ a,b,c,d$ be non-negative real numbers such that $ a+b+c+d=4$. Prove that $ 5(a^{2}+b^{2}+c^{2}+d^{2}) \ge a^{3}+b^{3}+c^{3}+d^{3}+16$. Also, the following is more interesting than the original one If $ a,b,c,d$ are positive and $ a+b+c+d=1$ then \[ 12(a^{3}+b^{3}+c^{3}+d^{3})+1\ge 7(a^{2}+b^{2}+c^{2}+d^{2}). \] I wait for a nice solution to it.

Vasc wrote: More interesting is the following: Let $ a,b,c,d$ be non-negative real numbers such that $ a+b+c+d=4$. Prove that $ 5(a^{2}+b^{2}+c^{2}+d^{2}) \ge a^{3}+b^{3}+c^{3}+d^{3}+16$. $ 5(a^{2}+b^{2}+c^{2}+d^{2}) \ge a^{3}+b^{3}+c^{3}+d^{3}+16\Leftrightarrow$ $ 5(a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d) \ge 4(a^{3}+b^{3}+c^{3}+d^{3})+(a+b+c+d)^{3}\Leftrightarrow$ $ \Leftrightarrow\sum_{sym}(a^{2}b-abc)\geq0,$ which true by Muirhead. hungkhtn wrote: If $ a,b,c,d$ are positive and $ a+b+c+d=1$ then \[ 12(a^{3}+b^{3}+c^{3}+d^{3})+1\ge 7(a^{2}+b^{2}+c^{2}+d^{2}). \] $ 12(a^{3}+b^{3}+c^{3}+d^{3})+1\ge 7(a^{2}+b^{2}+c^{2}+d^{2})\Leftrightarrow$ $ \Leftrightarrow12(a^{3}+b^{3}+c^{3}+d^{3})+(a+b+c+d)^{3}\ge 7(a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)\Leftrightarrow$ $ \Leftrightarrow\sum_{cyc}(a^{3}+b^{3}+c^{3}-a^{2}b-a^{2}c-b^{2}a-b^{2}c-c^{2}a-c^{2}b+3abc)\geq0.$

Yes,you are right. My inequality has equality when $ a=b=c=d$ or $ a=b=c,d=0$ and VASC inequality has equality when $ a=b=c=d$ or $ a=4,b=c=d=0.$

By Cauchy-Shwarz Inequality, (a^3+b^3+c^3+d^3)(a+b+c+d)=(a^3+b^3+c^3+d^3)>=(a^2+b^2+c^2+d^2)^2. So, 12(a^3+b^3+c^3+d^3)-7(a^2+b^2+c^2+d^2)+1>=12(a^2+b^2+c^2+d^2)^2-7(a^2+b^2+c^2+ +d^2)+1=(3(a^2+b^2+c^2+d^2)-1)(4(a^2+b^2+c^2+d^2)-1)=M. Left to prove, that m>=0. As, Average Square of a,b,c and d is more or equal to Average Arithmetical of a,b,c and d, a^2+b^2+c^2+d^2>=(a+b+c+d)^2/4=1/4. So, 4(a^2+b^2+c^2+d^2)-1>=0. So, we have to prove, that a^2+b^2+c^2+d^2>=1/3, but we can choose a,b,c and d so, that 1/4<a^2+b^2+c^2+d^2>1/3. For those numbers, the following inequality doesn't take place.

What do you mean? lasha? You mean my inequality is wrong? I feel something not true in your solution?!

hungkhtn wrote: What do you mean? lasha? You mean my inequality is wrong? I feel something not true in your solution?!

By AM-GM , have $6\sum a^3=\sum (2a^3+\frac{1}{8}a)+\sum (4a^3+\frac{1}{16}+\frac{1}{16})-\frac{5}{8}$ $\ge\sum a^2+\frac{3}{4}\sum a-\frac{5}{8}=\sum a^2+\frac{1}{8}.$

leepakhin wrote: Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$. Prove that\[ 6(a^3+b^3+c^3+d^3)\ge(a^2+b^2+c^2+d^2)+\frac{1}{8} \] Hong Kong 2006 France 2007 Turkey 2017 h

leepakhin wrote: Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$. Prove that\[ 6(a^3+b^3+c^3+d^3)\ge(a^2+b^2+c^2+d^2)+\frac{1}{8} \] Let $f(x)=6x^3-x^2-\frac1{32}$. The inequality is equivalent to $f(a)+f(b)+f(c)+f(d)\ge0$. Now by tangent-line trick, we have: $$f(x)-\frac58x+\frac5{32}=(4x-1)^2(3x+1)\ge0\Rightarrow f(x)\ge\frac58x-\frac5{32}$$and summing this over $a,b,c,d$ yields: $$f(a)+f(b)+f(c)+f(d)\ge\frac58(a+b+c+d)-\frac58=0$$as desired.