\(\dfrac{4}{3}\le a+\sqrt{ab}+\sqrt[3]{abc}=a+\sqrt[]{\dfrac{a}{2}.2b}+\sqrt[3]{\dfrac{a}{4}.b.4c}\)

\(\le a+\dfrac{1}{2}\left(\dfrac{a}{2}+2b\right)+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)=\dfrac{4}{3}\left(a+b+c\right)\)

\(\Rightarrow Q\ge1\)

\(Q_{min}=1\) khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)

Bài 1:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{2ab}+\frac{1}{a^2+b^2}\geq \frac{4}{2ab+a^2+b^2}=\frac{4}{a+b)^2}=4(1)\)

Áp dụng BĐT AM-GM:

\(1=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}\Rightarrow \frac{3}{2ab}\geq 6(2)\)

\(a^4+b^4\geq \frac{(a^2+b^2)^2}{2}\geq \frac{(\frac{(a+b)^2}{2})^2}{2}=\frac{1}{8}\) \(\Rightarrow \frac{a^4+b^4}{2}\geq \frac{1}{16}(3)\)

Từ \((1);(2);(3)\Rightarrow P\geq 4+6+\frac{1}{16}=\frac{161}{16}\)

Vậy \(P_{\min}=\frac{161}{16}\). Dấu bằng xảy ra tại $a=b=0,5$

Bài 2:
Áp dụng BĐT Cauchy-Schwarz:

\(2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)\geq 2. \frac{4}{x^2+y^2+2xy}=\frac{8}{(x+y)^2}=\frac{9}{2}\)

Áp dụng BĐT AM-GM:

\(\frac{80}{81xy}+5xy\geq 2\sqrt{\frac{80}{81}.5}=\frac{40}{9}\)

\(\frac{4}{3}=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{4}{9}\Rightarrow \frac{1}{81ab}\geq \frac{1}{36}\)

Cộng những BĐT vừa cm được ở trên với nhau:

\(\Rightarrow A\geq \frac{9}{2}+\frac{40}{9}+\frac{1}{36}=\frac{323}{36}\)

Vậy \(A_{\min}=\frac{323}{36}\Leftrightarrow a=b=\frac{2}{3}\)

Do \(a,b,c>0\) nên theo quy tắc phân số: \(\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\)

Tương tự: \(\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c}\); \(\dfrac{c}{a+c}< \dfrac{b+c}{a+b+c}\)

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Theo BĐT Cauchy: \(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\Leftrightarrow\dfrac{2}{a+b+c}\le\dfrac{1}{\sqrt{a\left(b+c\right)}}\)

\(\Leftrightarrow\dfrac{2a}{a+b+c}\le\sqrt{\dfrac{a}{b+c}}\)

Tương tự \(\dfrac{2b}{a+b+c}\le\sqrt{\dfrac{b}{a+c}}\); \(\dfrac{2c}{a+b+c}\le\sqrt{\dfrac{c}{a+b}}\)

(3 dấu = không thể đồng thời xảy ra, để chặt chẽ bạn có thể chia trường hợp)

Cộng vế với vế:

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}< \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)

\(\dfrac{4}{3}=a+2\sqrt{\dfrac{a}{4}.b}+\dfrac{1}{2}\sqrt[3]{\dfrac{a}{2}.2b.8c}\)

\(\dfrac{4}{3}\le a+\dfrac{a}{4}+b+\dfrac{1}{6}\left(\dfrac{a}{2}+2b+8c\right)=\dfrac{4}{3}\left(a+b+c\right)\)

\(\Rightarrow a+b+c\ge1\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)

Anh ơi cho em hỏi làm sao để tách/tìm điểm rơi như thế này ạ?

sai đề không ?

ko sai đề. mình vừa nghĩ ra rồi

P=\(a+\frac{1}{2}\sqrt{a.4b}+\frac{1}{4}\sqrt[3]{a.4b.16c} \le a+\frac {a+4b}{4}+\frac{a+4b+16c}{12}=\frac{4(a+b+c)}{3}=\frac{4}{3}\)

cái kia là \(3\sqrt{\dfrac{1}{a}+\dfrac{9}{b}+\dfrac{25}{c}}\)

\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)

\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)

\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)

\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)

\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)