Áp dụng BĐT Mincopxki:

\(P\ge\sqrt{\left(a+b+c\right)^2+2\left(a+b+c\right)^2}=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Lại có do \(a;b;c\ge0\) nên:

\(a^2+2b^2\le a^2+2\sqrt{2}ab+2b^2=\left(a+\sqrt{2}b\right)^2\)

\(\Rightarrow\sqrt{a^2+2b^2}\le a+\sqrt{2}b\)

Tương tự và cộng lại:

\(\Rightarrow P\le\left(\sqrt{2}+1\right)\left(a+b+c\right)=\sqrt{2}+1\)

Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(1;0;0\right)\) và các hoán vị

\(a;b\ge0\Rightarrow ab\ge0\)

\(\Rightarrow a^2+2b^2+2\sqrt{2}ab\ge a^2+2b^2\)

Áp dụng BĐT Bu-nhi-a-cốp-xki ta có:

\(S^2=\left(1.\sqrt{1+2a}+1.\sqrt{1+2b}\right)^2\le\left(1^2+1^2\right)\left(1+2a+1+2b\right)\)\(=4+4\left(a+b\right)\)

Áp dụng tiếp BĐT Bu-nhi-a-cốp-xki ta có \(\left(a+b\right)^2\le2\left(a^2+b^2\right)=2\)\(\Rightarrow a+b\le\sqrt{2}\)\(\Rightarrow S^2\le4+4\sqrt{2}\Rightarrow S\le2\sqrt{1+\sqrt{2}}\).Xảy ra đẳng thức khi và chỉ khi \(a=b=\frac{\sqrt{2}}{2}\)

Đề sai phải là \(\sqrt{2b^2+bc+2c^2}\)

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{5}{4}}\left(a+b\right)\)

CMTT, có: \(\sqrt{2b^2+bc+2c^2}\ge\sqrt{\frac{5}{4}}\left(b+c\right)\)

\(\sqrt{2c^2+ca+2a^2}\ge\sqrt{\frac{5}{4}}\left(c+a\right)\)

\(\Rightarrow P\ge\sqrt{5}\left(a+b+c\right)\ge\frac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\frac{\sqrt{5}}{3}\)

Dấu "=" xảy ra khi a=b=c=\(\frac{1}{9}\)

1) hệ <=> \(\left\{{}\begin{matrix}x+y+3\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)=1\\x+y+3\sqrt[3]{\left(x-1\right)\left(y+1\right)}\left(\sqrt[3]{x-1}+\sqrt[3]{y+1}\right)=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y+3\sqrt[3]{xy}=1\\x+y+3\sqrt[3]{\left(x-1\right)\left(y+1\right)}=1\end{matrix}\right.\)

trừ vế theo vế => \(3\sqrt[3]{xy}-3\sqrt[3]{\left(x-1\right)\left(y+1\right)}=0\)

<=> xy=(x-1)(y-1) <=> x-y=1=> \(\left\{{}\begin{matrix}\sqrt[3]{x}+\sqrt[3]{y}=1\\x-y=1\end{matrix}\right.\)

đặt \(\sqrt[3]{x}=a;\sqrt[3]{y}=b\)

hpt <=> \(\left\{{}\begin{matrix}a+b=1\\a^3-b^3=1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}b=1-a\\2a^3-3a^2+3a-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}b=1-a\\\left(a-1\right)\left(2a^2-a+2\right)=0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

p/s: cách làm khá dài ,có ai có cách khác thì làm luôn cho mik exp :v )

câu 2) xét \(p^2\) nhé ( để mai làm đã @@ )

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$P^2\leq (1+2a+1+2b)(1+1)=4(a+b+1)$

Tiếp tục áp dụng Bunhiacopxky:

$(a+b)^2\leq (a^2+b^2)(1+1)=2\Rightarrow a+b\leq \sqrt{2}$

$\Rightarrow P^2\leq 4(\sqrt{2}+1)$

$\Rightarrow P\leq 2\sqrt{\sqrt{2}+1}$

Vậy $P_{\max}=2\sqrt{\sqrt{2}+1}$. Giá trị này đạt tại $a=b=\frac{1}{\sqrt{2}}$

Đề thiếu nhé, a,b,c >0

Áp dụng BĐT Bunhiacopxki, ta có:

\(M^2=\left(\sqrt{2a+5\sqrt{ab}+2b}+\sqrt{2b+5\sqrt{bc}+2c}+\sqrt{2c+5\sqrt{ca}+2a}\right)^2\)

\(\le3\left[4\left(a+b+c\right)+5\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\right]\)

\(\le3\left[4\left(a+b+c\right)+5\left(a+b+c\right)\right]=81\)

\(\Rightarrow M\le9\)

\(MaxM=9\Leftrightarrow a=b=c=1\)

(\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\sqrt{\left(a+b+c\right)\left(a+b+c\right)}=a+b+c\left(Bunhiacopxki\right)\))