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![](https://rs.olm.vn/images/avt/0.png?1311)
P=\(\dfrac{\sqrt{2}.a}{\sqrt{\left(a^2+\left(b+c\right)^2\right)\left(1+1\right)}}+\dfrac{\sqrt{2}.b}{\sqrt{\left(b^2+\left(a+c\right)^2\right)\left(1+1\right)}}+\dfrac{\sqrt{2}.c}{\sqrt{\left(c^2+\left(b+a\right)^2\right)\left(1+1\right)}}\)>=\(\dfrac{\sqrt{2}.a}{\sqrt{\left(a+b+c\right)^2}}+\dfrac{\sqrt{2}.b}{\sqrt{\left(a+b+c\right)^2}}+\dfrac{\sqrt{2}.c}{\sqrt{\left(a+b+c\right)^2}}\)>=\(\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Theo bất đẳng thức Cô - si, ta có: \(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\le\frac{a+b+a+c}{2}+\frac{b+c}{2}=1\)
\(\Rightarrow\sqrt{a}\left(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\right)\le\sqrt{a}\)hay \(\sqrt{a^2+abc}+\sqrt{abc}\le\sqrt{a}\)
Tương tự ta có: \(\sqrt{b^2+abc}+\sqrt{abc}\le\sqrt{b}\);\(\sqrt{c^2+abc}+\sqrt{abc}\le\sqrt{c}\)
Mà \(abc\le\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}\Rightarrow\sqrt{abc}\le\frac{1}{3\sqrt{3}}\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\le3\left(a+b+c\right)=3\)\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\sqrt[3]{abc}\le a+b+c\Rightarrow abc\le\frac{\left(a+b+c\right)^3}{27}=\frac{1}{27}\) (BĐT AM-GM)
\(\sqrt{a^2+abc}=\sqrt{a\left(a+bc\right)}=\frac{2}{3}\sqrt{\frac{9}{4}a\left(a+bc\right)}\le\frac{2}{3}\left(\frac{\frac{9}{4}a+a+bc}{2}\right)\) (BĐT AM-GM)
Tương tự: \(\Rightarrow\)\(A\le\frac{1}{3}\left(\frac{9}{4}\left(a+b+c\right)+a+b+c+ab+bc+ca\right)+9\sqrt{\frac{1}{27}}\)
mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
=>giải được
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,b,c\ge0\Rightarrow abc\ge0\Rightarrow\sqrt{a^2+abc}\ge\sqrt{a^2}=a\)
Tương tự:\(\sqrt{b^2+abc}\ge b,\sqrt{c^2+abc}\ge c\)
\(\Rightarrow A\ge a+b+c+0=1\)
Đẳng thức xảy ra \(\Leftrightarrow abc=0,a+b+c=1\)(bạn tự giải tiếp)
0,3 0,4 ,0,5