\(c^2+d^2+25=6c+8d\)

\(\Leftrightarrow\left(c^2-6c+9\right)+\left(d^2-8d+16\right)=0\)

\(\Leftrightarrow\left(c-3\right)^2+\left(d-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}c-3=0\\d-4=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}c=3\\d=4\end{matrix}\right.\)

\(\Rightarrow P=25-3a-4b=25-\left(3a+4b\right)=25-Q\)

Xét \(Q=3a+4b\Rightarrow Q^2=\left(3a+4b\right)^2\le\left(3^2+4^2\right)\left(a^2+b^2\right)=25.2=50\)

\(\Rightarrow Q^2\le50\Rightarrow-5\sqrt{2}\le Q\le5\sqrt{2}\Rightarrow-Q\le5\sqrt{2}\)

\(\Rightarrow P\le25+5\sqrt{2}\)

\(P_{max}=25+5\sqrt{2}\) khi \(\left\{{}\begin{matrix}a^2+b^2=2\\\frac{a}{3}=\frac{b}{4}\\3a+4b=-5\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\frac{3\sqrt{2}}{5}\\b=-\frac{4\sqrt{2}}{5}\end{matrix}\right.\)

giúp mình bài này với ah.

\(3\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\Rightarrow a+b+c\le3\)

\(\Rightarrow\dfrac{a}{\sqrt[3]{3a+bc}}\le\dfrac{a}{\sqrt[3]{a\left(a+b+c\right)+bc}}=\sqrt[3]{2}.\sqrt[3]{\dfrac{a}{a+b}.\dfrac{a}{a+c}.\dfrac{a}{2}}\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{a}{2}\right)\)

Cộng vế và rút gọn:

\(E\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(E\le\dfrac{\sqrt[3]{2}}{3}\left(3+\dfrac{3}{2}\right)=\dfrac{3\sqrt[3]{2}}{2}\)

Lời giải:
Theo hệ quả quen thuộc của bđt AM-GM:
$(a+b+c)^2\leq 3(a^2+b^2+c^2)\leq 9$

$\Rightarrow a+b+c\leq 3$ (đpcm)

Từ đây ta có:

\(E\leq \frac{a}{\sqrt[3]{(a+b+c)a+bc}}+\frac{b}{\sqrt[3]{(a+b+c)b+ac}}+\frac{c}{\sqrt[3]{c(a+b+c)+ab}}\)

\(=\frac{a}{\sqrt[3]{(a+b)(a+c)}}+\frac{b}{\sqrt[3]{(b+c)(b+a)}}+\frac{c}{\sqrt[3]{(c+a)(c+b)}}\)

\(\leq \frac{\sqrt[3]{2}}{3}(\frac{a}{2}+\frac{a}{a+b}+\frac{a}{a+c})+\frac{\sqrt[3]{2}}{3}(\frac{b}{2}+\frac{b}{b+a}+\frac{b}{b+c})+\frac{\sqrt[3]{2}}{3}(\frac{c}{2}+\frac{c}{c+a}+\frac{c}{c+b})\)

\(=\frac{\sqrt[3]{2}(a+b+c)}{6}+\frac{\sqrt[3]{2}}{3}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})\leq \frac{3\sqrt[3]{2}}{2}\)

Vậy.................

ap dung bdt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) 

\(\frac{1}{2a+b+c}=\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)

\(\Rightarrow P\le\frac{1}{16}\left[\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2+\left(\frac{1}{a+b}+\frac{1}{b+c}\right)^2+\left(\frac{1}{b+c}+\frac{1}{a+c}^2\right)\right]\)

\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(a+c^2\right)}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(a+b\right)\left(a+c\right)}\)\(+\frac{2}{\left(b+c\right)\left(c+a\right)}\)

ap dung \(x^2+y^2+z^2\ge xy+yz+xz\) voi a+b=x, b+c=y, c+a=z

\(16P\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)

tiếp tục áp dụng bdt ban đầu \(\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Rightarrow\frac{1}{\left(a+b\right)^2}\le4.16.\left(\frac{1}{a}+\frac{1}{b}\right)^2\)

\(\Rightarrow16P\le\frac{1}{4}.16\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)

=\(\frac{1}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\right)\)

tiep tuc ap dung bo de thu 2 ta co 

\(16P\le\frac{1}{4}.4\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)

\(\Rightarrow p\le\frac{3}{16}\)dau =khi a=b=c=1

Nguồn : mạng :V vào thống kê coi hình

ÁP dụng BĐT bunhia có:

 \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\left(7-x\right)^2\le3\left(a^2+b^2+c^2\right)\) \(\Leftrightarrow-\dfrac{\left(7-x\right)^2}{3}\ge-\left(a^2+b^2+c^2\right)\)

Pt (2)\(\Leftrightarrow\)\(x^2=13-\left(a^2+b^2+c^2\right)\le13-\dfrac{\left(7-x\right)^2}{3}\)

\(\Leftrightarrow3x^2\le39-\left(7-x\right)^2\)

\(\Leftrightarrow4x^2-14x+10\le0\) \(\Leftrightarrow1\le x\le\dfrac{5}{2}\)

=>x_min=1 \(\Leftrightarrow\)a=b=c=2

x_max=\(\dfrac{5}{2}\)\(\Leftrightarrow\) a=b=c=\(\dfrac{3}{2}\)