Let $f(x)=6x^{3}-x^{2}$, we have $f'(x)=18x^{2}-2x$. Thus the tangent line of $y=f(x)$ at the point $(\frac{1}{4},\ f(\frac{1}{4})$ is $y=\frac{5}{8}(x-\frac{1}{4})+\frac{1}{32}\Longleftrightarrow y=\frac{5}{8}x-\frac{1}{8}$.Then for $x>0$, $f(x)-(\frac{5}{8}x-\frac{1}{8})=6(x-\frac{1}{4})^{2}(x+\frac{1}{3})\geq 0$, yielding $6x^{3}-x^{2}\geq \frac{5}{8}x-\frac{1}{8}$. Since $a,\ b,\ c,\ d>0$ and $a+b+c+d=1$, $\therefore 6(a^{3}+b^{3}+c^{3}+d^{3})-(a^{2}+b^{2}+c^{2}+d^{2})\geq \frac{5}{8}(a+b+c+d)-\frac{1}{8}\cdot 4=\frac{1}{8}$. Q.E.D.

Igor wrote: Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}. \] Alternatively, it is obvious according to SID theorem since \[f(1,0,0,0) \ge 0 ; f(\frac{1}{2},\frac{1}{2},0,0) \ge 0 ; f(\frac{1}{3},\frac{1}{3},\frac{1}{3},0) \ge 0 , f(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}) \ge 0,\] in which \[f(x)=6(a^{3}+b^{3}+c^{3}+d^{3})-a^{2}-b^{2}-c^{2}-d^{2}-\frac{1}{8}.\]

My solution was the following : by using chebyshev : $4(a^{3}+b^{3}+c^{3}+d^{3}) \geq (a+b+c+d)(a^{2}+b^{2}+c^{2}+d^{2})$ so $4(a^{3}+b^{3}+c^{3}+d^{3}) \geq a^{2}+b^{2}+c^{2}+d^{2}$ and we have $2(a^{3}+b^{3}+c^{3}+d^{3}) \geq \frac{(a^{2}+b^{2}+c^{2}+d^{2})}{2}\geq \frac{(a+b+c+d)^{2}}{8}=\frac{1}{8}$ we are done

Nice solution. Very simple.

I went to Turkey Olympiad Studying Winter Camp in February. This question was asked to us in camp finish exam.

i can't believe this is a TST problem. from chebyshev we have \[4(a^{3}+b^{3}+c^{3}+d^{3})\ge (a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)\]and from power mean \[\frac{a^{3}+b^{3}+c^{3}+d^{3}}{4}\ge \left(\frac{a+b+c+d}{4}\right)^{3}\]that is \[2(a^{3}+b^{3}+c^{3}+d^{3})\ge \frac18\]adding we get the result.

A doubt please explain Tchebychev's inequality can be applied only if $ a_1\ge a_2\ge .......\ge a_n$ and similarly for $ b_n$. Then how come you have applied directly?

Aravind Srinivas L wrote: A doubt please explain Tchebychev's inequality can be applied only if $ a_1\ge a_2\ge .......\ge a_n$ and similarly for $ b_n$. Then how come you have applied directly? Now I'm not sure about the Tchebychev inequality. But i think it is something like $ (a_1 + a_2 + ... + a_n)(b_1 + b_2 + ... + b_n) \le n(a_1b_1 + a_2b_2 + ... + a_nb_n)$ If i'm correct it doesn't matter if $ a_1 \ge a_2 \ge .. \ge a_n$. We just have to have $ a$ similar sorted as $ b$.

What does SID theorem stand for?

Symmetric inequality of Degree 3 theorem - SID theorem see http://www.mathlinks.ro/viewtopic.php?search_id=597762145&t=148928

Aravind Srinivas L wrote: A doubt please explain Tchebychev's inequality can be applied only if $ a_1\ge a_2\ge .......\ge a_n$ and similarly for $ b_n$. Then how come you have applied directly? Chebyshev's inequality works here because $ (a^2,b^2,c^2,d^2)$ and $ (a,b,c,d)$ are always sorted similarly. In fact you can assume $ a\ge b\ge c\ge d$ due to symmetry and verify the fact.

this problem can be solved easily with ilham's cauchy.

$ 48(a^3 + b^3 + c^3 + d^3) \ge 8(a^2 + b^2 + c^2 + d^2)(a + b + c + d) + (a + b + c + d)^3$ so we are done by muirhead

mm.. maybe i have another solution by the cauchy-schwarz inequality $(a^3+b^3+c^3+d^3)(a+b+c+d) \geq (a^2+b^2+c^2+d^2)^2$ hence $(a^3+b^3+c^3+d^3) \geq (a^2+b^2+c^2+d^2)^2$ now we set the $LHS=x$ and $RHS=y$ and we have a new system of inequalities $x\geq y^2$ $6x\geq y+ 1/8$ $\Rightarrow 6x\geq 6y^2 \geq y+1/8$ thus we have $6y^2-y-1/8 \geq 0 $ from wich follows $y\leq -1/12 , y \geq 1/4$ since $y$ is always positive we just have to show that $y \geq 1/4$ wich follows from cauchy schwarz : $ 4(a^2+b^2+c^2+d^2) \geq (a+b+c+d)=1$ , can someone check if it's correct ? it's a new topic for me, thanks!!

That's a nice solution, dudeldai.

Using $a+b+c+d=1$, we get $156S(3,0,0) \geq 120S(2,1,0)+24S(1,1,1)$ Obviously this holds by Muirhead inequality

Igor wrote: Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}.\] https://artofproblemsolving.com/community/c6h1531976p9218816

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.

dudeldai wrote: $x\geq y^2$ $6x\geq y+ 1/8$ $\Rightarrow 6x\geq 6y^2 \geq y+1/8$ thus we have $6y^2-y-1/8 \geq 0 $ If $a \geq b$ and $a \geq c$, this does not say anything about whether $b \geq c$...

CAUCHY $(a^3+b^3+c^3+d^3)>=\frac{(a^2+b^2+c^2+d^2)^2}{a+b+c+d}$ $(a^3+b^3+c^3+d^3)>={(a^2+b^2+c^2+d^2)^2}>=\frac{a^2+b^2+c^2+d^2}{4}$ CAUCHY $a^2+b^2+c^2+d^2>= \frac{1}{4}$ $6(a^3+b^3+c^3+d^3)>=\frac{3}{2}*(a^2+b^2+c^2+d^2)>=a^2+b^2+c^2+d^2+\frac{1}{8}$

This can be solved just by using Chebyshev's inequality 2x lol

hmm guys, what is SID Theorem??

eeshift wrote: Symmetric inequality of Degree 3 theorem - SID theorem see http://www.mathlinks.ro/viewtopic.php?search_id=597762145&t=148928 ^^

but it said "file not found" ):

Igor wrote: Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}.\] $$8LHS-8RHS=64\sum a^3 -8\sum a^2 \sum a -(\sum a)^3=11\sum (a-b)(a-c)(a-d) +\frac{28\sum (a^3+b^3+c^3-3abc)}{3} \geq 0$$

Let $f(a,b,c,d)=6a^3+6b^3+6c^3+6d^3-a^2-b^2-c^2-d^2\text{ and }g(a,b,c,d)=a+b+c+d-2$, furthermore, define $L=f(a,b,c,d)-\lambda g(a,b,c,d)$ $$\frac{\partial L}{\partial a}=18a^2-2a-\lambda=0$$$$\frac{\partial L}{\partial b}=18b^2-2a-\lambda=0$$$$\frac{\partial L}{\partial c}=18c^2-2c-\lambda=0$$$$\frac{\partial L}{\partial d}=18d^2-2d-\lambda=0$$Thus $18a^2-2a=18b^2-2b=18c^2-2c=18d^2-2d\Longrightarrow 9a^2-a=9b^2-b=9c^2-c=9d^2-d$ and let (1),(2),(3) and (4) denote the equations respectively. Furthermore from (1) and (2) we obtain $9a^2-a=9b^2-b\Longrightarrow 9(a^2-b^2)=a-b\Longrightarrow 9(a+b)(a-b)=a-b$ thus $9(a+b)=1\Longleftrightarrow a+b=\frac{1}{9}$ if and only if $a\neq b$ Also, from (3) and (4) we obtain $9c^2-c=9d^2-d\Longrightarrow 9(c+d)(c-d)=c+d$ thus $9(c+d)=1\Longleftrightarrow c+d=\frac{1}{9}$ if and only if $c\neq d$ Therefore, if $a\neq b\text{ and }c\neq d$, we get that $a+b+c+d=\frac{1}{9}+\frac{1}{9}=\frac{2}{9}<1$ which is clearly a contradiction, thus $a=b\text{ and }c=d$ (we obtain the same contradiction from the other equations) Since we can obtain the same contradiction from any other permutation, we can conclude that $a=b=c=d$, thus $4a=1\Longrightarrow a=b=c=d=\frac{1}{4}$ Therefore the minimum of the function is achieved at $f\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)=\frac{1}{8}$ $\blacksquare$.

$6(a^3+b^3+c^3+d^3) \geq 6(a^2+b^2+c^2+d^2)^2$ $a^2+b^2+c^2+d^2 \geq \frac{1}{4}$ Call $x=a^2+b^2+c^2+d^2$ Then $12x^2 \geq 3x$ And we are done...

Igor wrote: Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}.\] Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[12(a^{3}+b^{3}+c^{3}+d^{3})\geq 7(a^{2}+b^{2}+c^{2}+d^{2})-1.\]

Anyone? Very easy

mihaig wrote: Igor wrote: Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}.\] Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$. Prove that: \[12(a^{3}+b^{3}+c^{3}+d^{3})\geq 7(a^{2}+b^{2}+c^{2}+d^{2})-1.\] Pretty easy as you said. $12(a^3+b^3+c^3+d^3)=12(\frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}+\frac{d^4}{d})\geq 12(\frac{(a^2+b^2+c^2+d^2)^2}{a+b+c+d})=12(a^2+b^2+c^2+d^2)^2$ $(a^2+b^2+c^2+d^2)(12(a^2+b^2+c^2+d^2)-7)+a+b+c+d\geq 0$ $\frac{(a+b+c+d)^2}{4}\left(3(a+b+c+d)^2-7)\right)+a+b+c+d\geq 0$ $3x^4-7x^2\geq -4$ Which is true.