\(A=\frac{a\sqrt{a}}{\sqrt{a+b+2c}}+\frac{b\sqrt{b}}{\sqrt{b+c+2a}}+\frac{c\sqrt{c}}{\sqrt{c+a+2b}}\)

\(A=\frac{a^2}{\sqrt{a\left(a+b+2c\right)}}+\frac{b^2}{\sqrt{b\left(b+c+2a\right)}}+\frac{c^2}{\sqrt{c\left(c+a+2b\right)}}\)

\(\ge\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\)

Xét: \(2\left(\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\right)\)

\(=\sqrt{4a\left(a+b+2c\right)}+\sqrt{4b\left(b+c+2a\right)}+\sqrt{4c\left(c+a+2b\right)}\)

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

\(\Rightarrow\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\le2\left(a+b+c\right)\)

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

\("="\Leftrightarrow a=b=c=1\)

123

ai tích mk lên 885 mk tích lại cho 

Đề bài có vấn đề nho nhỏ, thay điểm rơi vào thì vế phải thừa bình phương trong ngoặc

Áp dụng Holder:

\(\left(a^2+\frac{1}{b^2}\right)\left(4+\frac{1}{4}\right)\left(4+\frac{1}{4}\right)\ge\left(\sqrt[3]{16a^2}+\sqrt[3]{\frac{1}{16b^2}}\right)^3\)

\(\Rightarrow\sqrt[3]{17^2\left(a^2+\frac{1}{b^2}\right)}\ge4\sqrt[3]{4a^2}+\frac{1}{\sqrt[3]{b^2}}\)

\(\Rightarrow P=\sqrt[3]{17^2}.S\ge4\sqrt[3]{4}\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}\right)+\frac{1}{\sqrt[3]{a^2}}+\frac{1}{\sqrt[3]{b^2}}+\frac{1}{\sqrt[3]{c^2}}\)

\(P=\frac{15}{\sqrt[3]{16}}\sum\sqrt[3]{a^2}+\sum\left(\sqrt[3]{\frac{a^2}{16}}+\frac{1}{\sqrt[3]{a^2}}\right)\)

Ta có: \(3\sqrt[3]{a^2}+\sqrt[3]{4}\ge4\sqrt[12]{4a^6}=4\sqrt[6]{2}.\sqrt{a}\)

Tương tự và cộng lại:

\(\Rightarrow\sum\sqrt[3]{a^2}\ge\frac{4\sqrt[6]{2}\sum\sqrt{a}-3\sqrt[3]{4}}{3}\ge3\sqrt[3]{4}\)

\(\sum\left(\sqrt[3]{\frac{a^2}{16}}+\frac{1}{\sqrt[3]{a^2}}\right)\ge6\sqrt[6]{\frac{1}{16}}=\frac{6}{\sqrt[3]{4}}\)

\(\Rightarrow P\ge\frac{15}{\sqrt[3]{16}}.3\sqrt[3]{4}+\frac{6}{\sqrt[3]{4}}=\frac{51}{\sqrt[3]{4}}=3.\sqrt[3]{\frac{17^3}{4}}\)

\(\Rightarrow S\ge3\sqrt[3]{\frac{17^3}{4}}:\sqrt[3]{17^2}=3\sqrt[3]{\frac{17}{4}}\)

Dấu "=" xảy ra khi \(a=b=c=2\)

Bài toán nhạt nhẽo, chẳng có gì ngoài tính trâu, lần sau xin né :(

Áp dụng bất đẳng thức Cauchy Schwarz thôi bạn

\(S\ge3\frac{1}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right)}}.\)
\(S\ge\frac{3.4}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right).16.16.16}}\)

\(S\ge\frac{12}{\frac{a+2b+5c+b+2c+5a+c+2a+5b+16+16+16}{6}}\)

\(S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=2\)

1.

\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)

\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)

Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá

2.

\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)

Đặt \(x+y+z=t\Rightarrow0< t\le1\)

\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

3.

\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)

Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)

Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)

4.

ĐKXĐ: \(-2\le x\le2\)

\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)

\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)

Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)

\(y_{min}=-2\) khi \(x=-2\)

5.

\(\frac{a\sqrt{b-1}+b\sqrt{a-1}}{ab}=\frac{1.\sqrt{b-1}}{b}+\frac{1.\sqrt{a-1}}{a}\le\frac{1+b-1}{2b}+\frac{1+a-1}{2a}=1\)

\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le ab\)

Dấu "=" xảy ra khi \(a=b=2\)

6. Áp dụng BĐT cơ bản:

\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow\left(ab+bc+ca\right)^2\ge3\left(ab.bc+bc.ca+ab+ca\right)\)

\(\Rightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Dấu "=" xảy ra khi \(a=b=c\)