The pqr handout 习题求助

hbghlyj

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下面开始习题：
$p = a + b, q = ab.$
The Basic Theory (Two variables)有2道习题：

• Which conditions (particularly, inequalities) should satisfy $p$ and $q$ for $a$ and $b$ to be real?
• Prove that only if $p$ and $q$ are non-negative real numbers which satisfy conditions from the previous problem then $a$ and $b$ are real and non-negative.

题1就是$p^2\ge4q$
题2就是由$p^2\ge4q$推出$a,b\inR$，然后容易由$a+b\ge0,ab\ge0$推出$a\ge0,b\ge0$.

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1.5 The abc Method有3道题：
3.1. Let $a,b,c$ be positive reals such that $a+b\geq c,\:b+c\geq a$ and $c+a\geq b$. Prove that
$$
2a^2(b+c)+2b^2(c+a)+2c^2(a+b)\geq a^3+b^3+c^3+9abc
$$
Hint: Use the incircle substitution including the degenerate case.

3.1. 代换$x=b+c-a,y=c+a-b,z=a+b-c$
\begin{multline*}2a^2(b+c)+2b^2(c+a)+2c^2(a+b)-(a^3+b^3+c^3+9abc)\\=\frac14 (x^3+ y^3+ z^3 - x^2 y - x^2 z - x y^2 - x z^2  - y^2 z - y z^2 + 3 x y z )\ge0\end{multline*}即为Schurs Inequality $n=3$.

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3.2.1. Express $s^k$ for $1\leq k\leq6$ in terms of $p,q,r$. You can check your expressions in Shortcuts
Hint: Express $(a+b+c),(a+b+c)^{2},(a+b+c)^{3}$, then can you continue?
\begin{gathered}\sum x=p \\ \sum x^2=p^2-2 q \\ \sum x^3=p^3-3 p q+3 r\end{gathered}

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3.2.2. Let $a,b,c$ be real numbers such that

$a+b+c=9$
$ab+bc+ca=24$

Prove that $16\leq abc\leq20$.
Prove moreover that for any $r\in[16,20]$ there exist real numbers $a,b,c$ such that

$a+b+c=9$
$ab+bc+ca=24$
$abc =r$

Hint: Substitute the values in $T(p,q,r)$.

3.2.2. 将$p=9,q=24$代入Theorem 1.3.3 $r \in\left[\frac{9 p q-2 p^3-2 \sqrt{\left(p^2-3 q\right)^3}}{27}, \frac{9 p q-2 p^3+2 \sqrt{\left(p^2-3 q\right)^3}}{27}\right]$即可

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3. Suppose $a,b,c$ are positive real numbers such that $abc=1$ and
\begin{aligned}\frac1a+\frac1b+\frac1c&=1=\frac1{ab}+\frac1{bc}+\frac1{ca}\end{aligned}Find the minimum value of $(a+1)(b+1)(c+1)$
Hint: Clear denominators.

3. $p=q=r=1\implies a,b,c$是方程$x^3-x^2+x-1=0$的根$\implies a=b=c=1,(a+1)(b+1)(c+1)=8$.岂不太简单了?

hbghlyj

第10页习题是证明这些不等式
\begin{gather}
p q \geq 9 r \\
p^2 \geq 3 q \\
q^2 \geq 3 p r \\
p^2 q+3 p r \geq 4 q^2 \\
p q^2 \geq 2 p^2 r+3 q r \\
p^2 q^2+12 r^2 \geq 4 p^3 r+p q r \\
p^3 \geq 27 r \\
q^3 \geq 27 r^2 \\
p^3 r \geq q^3 \\
p^3+9 r \geq 4 p q\label{10} \\
2 p^3+9 r \geq 7 p q \\
2 p^3+27 r \geq 9 p q \\
2 p^3+9 r^2 \geq 7 p q r \\
q^3+9 r^2 \geq 4 p q r \\
2 q^3+27 r^2 \geq 9 p q r \\
p^4+3 q^2 \geq 4 p^2 q \\
p^4+4 q^2+6 p r \geq 5 p^2 q \\
4 p^5 q+44 p^2 q r+17 p q^3 \geq 4 p^4 r+20 p^3 q^2+25 q^2 r+24 p r^2
\end{gather}

hbghlyj

本论坛无法搜索$+$号
搜不了以前的帖子

hbghlyj

第14页习题

• Let $a, b, c$ be non-negative real numbers such that $a+b+c=1$. Show that
  $$
  1+12 a b c \geq 4(a b+b c+c a)
  $$
• Can you similarly prove the q-lemma and r-lemma?
  Hint: Cubic polynomial in $q$ then consider then set of non-negative values of $x$ for which $g(x)$ is non-negative, in this case $q=0$. Do the same for $r \neq 0$, but over here leading term is negative, do a similar argument, in this case $u=0$.

hbghlyj

hbghlyj 发表于 2023-8-5 23:31
第14页习题

• Let $a, b, c$ be non-negative real numbers such that $a+b+c=1$. Show that
  

题1齐次化是$p^3+12r\ge4pq$，条件为$a,b,c\inR_{\ge0}$
\eqref{10}是$p^3+9r\ge4pq$，条件为$a,b,c\inR$
好像用\eqref{10}能推出题1

O-17

先解决个看起来最吓人的 $\color{blue}{(18)}$ 吧, 将 $p,q,r$ 代换展开之后等价于证明
$$
\sum\left(4a^6b+4a^6c-3a^4b^3-a^4b^2c-a^4bc^2-3a^4c^3-2a^3b^3c+2a^3b^2c^2\right)\geqslant0.
$$
其实是非常简单的, 外圈可以用若干个 $\left[1,-1,-1,1\right]$ 结构消掉, 剩下的可以拆成一个四次 Schur 和一个均值, 将上式的左边简记作 $u$ , 注意到
$$
u=3\sum ab\left(a^2-b^2\right)\left(a^3-b^3\right)+\sum c\left(a-b\right)\left(a^5-b^5\right)+abc\left[2\sum a^2(a-b)(a-c)+\sum ab(a-b)^2\right]
$$
然后再把上面每一项通通写成 $f(a,b,c)\cdot(a-b)^2$ 的形式,
\begin{align*}
\left(a^2-b^2\right)\left(a^3-b^3\right)={}&(a+b)\left(a^2+ab+b^2\right)(a-b)^2\\
\left(a-b\right)\left(a^5-b^5\right)={}&\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)(a-b)^2\\
2\sum a^2(a-b)(a-c)={}&\sum(a+b-c)^2(a-b)^2\\
\end{align*}
整理一下, 你就会得到本题的显式 SOS 结果,
$$
u=\sum X(a-b)^2
$$
其中
\begin{align*}
X:={}&3ab(a+b)\left(a^2+ab+b^2\right)+c\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\\
&+abc(a+b-c)^2+a^2b^2c\\
\geqslant{}&0.~\square
\end{align*}
能显式 SOS 就足以说明这个不等式并不强了, 七次里面比它强的结构一抓一大把, 也不知道为啥偏偏拿这个当习题, 这感觉也 "Handy" 不到哪去啊...