by AM-GM, we have $\sum a\geq 3$ and $\sum a^2 =\sum (a-1)^2+2\sum a -3\geq \sum a$. so, \begin{align*}p\sum a^2+q\sum \frac{1}{a}&=p\sum a^2+q\sum ab\\&=(p-q)\sum a^2+q\sum (a^2+ab)\\&=(p-q)\sum a^2+\frac{1}{2}q\sum (a+b)^2\\&\geq (p-q)\sum a+\frac{1}{2}q(\frac{4(\sum a)^2}{3})\\&=(p-q)\sum a+2q\sum a =(p+q)\sum a\end{align*} cmiiw

$ p(a^2+b^2+c^2)+q\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=p(a^2+b^2+c^2)+q(ab+ac+bc)=\frac{p+q}{3} (a^2+b^2+c^2)+\frac{2p-q}{3}(a^2+b^2+c^2)+q(ab+ac+bc)\geq \frac{p+q}{3}(a^2+b^2+c^2)+\left(\frac{2p-q}{3}+q\right)(ab+ac+bc)=\frac{p+q}{3} (a+b+c)^2=(p+q)(a+b+c) \frac{a+b+c}{3}\geq (p+q)(a+b+c) $

Is known that:(My Inequality Project Ramchandran,pb.15 ) If k is a real number and $ a,b,c>0,abc=1\Rightarrow\sum{a^2}+k\sum{ab}\ge (k+1)\sum{a} $ For $ q>0,k=\frac{p}{q} $ we obtain the inequality proposed! _____ Marin Sandu

hello, your link doesn't work. Sonnhard.

Here is a slightly different solution: By AM-GM, we have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 3.$ Therefore, by AM-GM again, we have \[a^2 + b^2 + c^2 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \left(a^2 + 1\right) + \left(b^2 + 1\right) + \left(c^2 + 1\right) \ge 2(a + b + c).\] In addition, since $1 + 1 + 1 \le a + b + c$ by AM-GM, we can strengthen the last inequality above into \[a^2 + b^2 + c^2 \ge a + b + c.\] Then since $p - q \ge 0$, we can weight these two inequalities as follows: \begin{align*} q\left(a^2 + b^2 + c^2 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) &\ge q\left(2(a + b + c)\right) \\ (p - q)\left(a^2 + b^2 + c^2\right) &\ge (p - q)(a + b + c). \end{align*} Summation yields the desired result. $\square$

my solution which is similar to Dukejuke's solution= by AM-GM $2a^2+\frac{1}{a}\ge 3a$ so $3(a^2+b^2+c^2)\ge 2(a^2+b^2+c^2)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 2(a+b+c)$ or $q(a^2+b^2+c^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 2q(a+b+c)$ = (2) similarly $3(a^2+b^2+c^2)\ge 2(a^2+b^2+c^2+ab+bc+ca)\ge 3(a+b+c)$ as $abc=1$ or $a^2+b^2+c^2\ge a+b+c$ and since $p\ge q$ we have $(p-q)(a^2+b^2+c^2)\ge (p-q)(a+b+c)$ =(1) adding (1) and (2) yields $ p \left(a^2+b^2+c^2\right) + q\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq (p+q) (a+b+c).$ so we are done

Given positive reals $a,b,c,p,q$ satisfying $abc=1$ and $p \geq \frac{q}{2}$ , prove that \[ p \left(a^2+b^2+c^2\right) + q\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq (p+q) (a+b+c). \]

sqing wrote: Given positive reals $a,b,c,p,q$ satisfying $abc=1$ and $p \geq \frac{q}{2}$ , prove that \[ p \left(a^2+b^2+c^2\right) + q\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq (p+q) (a+b+c). \] $\implies p(\sum a^2-\sum a)\ge q(\sum a-\sum \frac{1}{a})$ so it remains to prove the stronger inequality $\sum a^2-\sum a\ge 2(\sum a-\sum \frac{1}{a})\implies \sum a^2+2\sum \dfrac{1}{a}\ge 3\sum a$ However, this is true by summing this AM-GM cyclically: $$a^2+\dfrac{1}{b}+\dfrac{1}{c}\ge 3\sqrt[3]{\dfrac{a^2}{bc}}=3a$$

The result $a^2 + b^2 + c^2 \ge ab + bc + ca$ implies stronger result $$a^2 + b^2 + c^2 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{2}{3} (a^2 + b^2 + c^2) + \frac{4}{3} (ab + bc + ca)$$$$= \frac{2}{3} \sum_{cyc} (a^2 + ab + ac) \ge \frac{2}{3} \cdot 3 \sum_{cyc} a^{4/3} b^{1/3} c^{1/3} = 2(a + b + c).$$(First inequality proves original problem, second inequality proves second problem)

What's the largest constant $k$ such that the inequality is true with $p\ge \dfrac{q}{k}$? I found $k\approx 15.234422+$ but that seems very absurd...

Solution using Tchebychev inequality : we have : $ p \ge q $ and $ a^2+b^2+c^2 \ge ab+ac+bc = \frac{1}{a} + \frac{1}{b}+ \frac {1}{c} $ so Tchebychev implies : $ p(a^2+b^2+c^2) +q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c} ) \ge 1/2 (p+q)(a^2+b^2+c^2+\frac{1}{a} + \frac{1}{b}+ \frac {1}{c} ) \ge (p+q)(a+b+c)$ since abc=1

john10 wrote: Solution using Tchebychev inequality : we have : $ p \ge q $ and $ a^2+b^2+c^2 \ge ab+ac+bc = \frac{1}{a} + \frac{1}{b}+ \frac {1}{c} $ so Tchebychev implies : $ p(a^2+b^2+c^2) +q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c} ) \ge 1/2 (p+q)(a^2+b^2+c^2+\frac{1}{a} + \frac{1}{b}+ \frac {1}{c} ) \ge (p+q)(a+b+c)$ since abc=1 I don't understand one thing that why we also make the problem to be difficult when it realy seems to be easy: Since $ a^2+b^2+c^2 \ge ab+ac+bc = \frac{1}{a} + \frac{1}{b}+ \frac {1}{c} $ we get $p(a^2+b^2+c^2)+q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c}) \ge p(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c})+q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c})=RHS$. Not using Tchebychev inequality or any strong.

Re1gnover wrote: john10 wrote: Solution using Tchebychev inequality : we have : $ p \ge q $ and $ a^2+b^2+c^2 \ge ab+ac+bc = \frac{1}{a} + \frac{1}{b}+ \frac {1}{c} $ so Tchebychev implies : $ p(a^2+b^2+c^2) +q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c} ) \ge 1/2 (p+q)(a^2+b^2+c^2+\frac{1}{a} + \frac{1}{b}+ \frac {1}{c} ) \ge (p+q)(a+b+c)$ since abc=1 I don't understand one thing that why we also make the problem to be difficult when it realy seems to be easy: Since $ a^2+b^2+c^2 \ge ab+ac+bc = \frac{1}{a} + \frac{1}{b}+ \frac {1}{c} $ we get $p(a^2+b^2+c^2)+q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c}) \ge p(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c})+q(\frac{1}{a} + \frac{1}{b}+ \frac {1}{c})=RHS$. Not using Tchebychev inequality or any strong. False!

Generalization: Suppose $k=5+6\sqrt{3}\cos \frac{\pi}{18}$, and $p\geq \frac{q}{k}$. Prove that \[ p \left(a^2+b^2+c^2\right) + q\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq (p+q) (a+b+c). \]

bobthesmartypants wrote: I found $k\approx 15.234422+$ but that seems very absurd... That is $k=5+6\sqrt{3}\cos \frac{\pi}{18}$, indeed. Occurs on $a:b:c=8\cos^3 \frac{\pi}{9}:1:1$.

See also here: http://artofproblemsolving.com/community/c6h36874p827426

Since $3(a^2+b^2+c^2) \ge (a+b+c)^2 \ge 3(a+b+c)$ by AM-GM, we have $a^2+b^2+c^2 \ge a+b+c$. Now we prove $\sum_{cyc} (a^2+\frac{1}{a}) \ge \sum_{cyc} 2a$. This is easy since $\sum_{cyc} (a^2-2a) \ge -3$ by Trivial Inequality and $\sum_{cyc} \frac{1}{a} \ge 3$ by AM-GM. $\blacksquare$. This gives us $$q(a^2+b^2+c^2) + q(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 2q(a+b+c)$$and adding this with $$(p-q)(a^2+b^2+c^2) \ge (p-q)(a+b+c)$$gives the desired inequality. $\blacksquare$

correct me if im wrong plz but doesnt writing $p=q+r$ end the problem since we know $a^2+b^2+c^2$ is bigger than $a+b+c$?

No, it doesn’t.

WolfusA wrote: No, it doesn’t. ah yes it does as you can see in @rkm0959's solution.

But he didn’t do just a simple substitution. If your answer is yes, then why do you ask?

WolfusA wrote: But he didn’t do just a simple substitution. If your answer is yes, then why do you ask? honestly it was simple after that but sorry I was harsh it just seemed wrong to me because this is an A2 after all. i expected sth a bit harder. in my defense i did mention that $a^2+b^2+c^2$ is bigger than $a+b+c$ and substituting p=q is just simple AM-GM anyways thanks and sorry

By weighted Am-Gm, $\frac{p}{p+q}(a^2+b^2+c^2)+\frac{q}{p+q}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge (a^2+b^2+c^2)^ \frac{p}{p+q}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^ {\frac{q}{p+q}}$ Now by c-s,$(a^2+b^2+c^2)\ge \frac{(a+b+c)^2}3$ and $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 3$ So the last inequality becomes $(a^2+b^2+c^2)^\frac{p}{p+q}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^\frac{q}{p+q}\ge \frac{(a+b+c)^\frac{2p}{p+q}}{3^\frac{p}{p+q}}3^\frac{q}{p+q}=(a+b+c)^\frac{2p}{p+q}(1/3)^\frac{p-q}{p+q}\ge (a+b+c)$ The last inequality follows from the fact that $p\ge q$ so $\frac{2p}{p+q}\ge 1$

Let $a+b+c = x$ and $ab+bc+ca = y$ and $abc = z =1$. $x^2 \ge 3y , y^2 \ge 3xz = 3x \implies x \ge 3 \implies y \ge 3$ we need to prove $px^2 - 2py + q((\sum \frac{1}{a})z) = px^2 - 2py + qy \ge (p+q)x$ $ px^2 - 2py + qy \ge px^2 - \frac{2}{3}px^2 + 3q = \frac{p}{3}x^2 + 3q = p(\frac{1}{3}x^2) + q(3) \ge (p+q)\sqrt[p+q]{\frac{x^{2p}}{3^{q-p}}}$ and since $x \ge 3$ we have $\frac{x^{2p}}{3^{q-p}} \ge x^{p+q}$ so $ px^2 - 2py + qy \ge (p+q)\sqrt[p+q]{\frac{x^{2p}}{3^{q-p}}} \ge (p+q)\sqrt[p+q]{x^{p+q}} = (p+q)x$

Let $a+b+c=3u, \ ab+bc+ca=3v^2$. We have $u\geq v\geq 1$. \[p(9u^2-6v^2)+q(3v^2)\overset{?}{\geq} (p+q)(3u)\iff p(3u^2-2v^2)+qv^2\overset{?}{\geq}(p+q)u\]\[3u^2p-2v^2p+qv^2\overset{?}{\geq} pu+qu\iff u(3up-p-q)\overset{?}{\geq} v^2(2p-q)\]Since $3up-p-q\geq 3p-p-q=2p-q\geq p>0$ and $2p-q\geq p>0,u\geq v\geq 1,$ \[u(3up-p-q)\geq v(3vp-p-q)\geq v(2vp-q)\geq v(2vp-vq)=v^2(2p-q)\]As desired.$\blacksquare$