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

\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2a}{2b+a+b}+\dfrac{2b}{2c+b+c}+\dfrac{2c}{2a+a+c}\)

\(\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)=2\left(\dfrac{a^2}{a^2+3ab}+\dfrac{b^2}{b^2+3bc}+\dfrac{c^2}{c^2+3ca}\right)\)

\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)

\(\Rightarrow P\ge\dfrac{3\sqrt{2}}{2}\) (đpcm)

\(\dfrac{a}{\sqrt{ab+b^2}}=\dfrac{\sqrt{2}.a}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{\sqrt{2}.a}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)

làm tương tự với \(\dfrac{b}{\sqrt{bc+c^2}};\dfrac{c}{\sqrt{ca+a^2}}\)

\(=>P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\)

\(=2\sqrt{2}\left(\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)

\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+\dfrac{4}{3}\left(ab+bc+ca\right)+\dfrac{8}{3}\left(ab+bc+ca\right)}\right]\)

\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{\dfrac{4}{3}\left(a+b+c\right)^2}\right]=\dfrac{2\sqrt{2}.3}{4}=\dfrac{3\sqrt{2}}{2}\)

dấu"=" xảy ra<=>a=b=c

e)

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

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)

=> ĐPCM

BPT?

Áp dụng bđt cosi schwart ta có:

`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`

Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`

`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`

Dấu "=" `<=>a=b=c=1.`

1) \(\left(a-b\right)^2\ge0\)

\(a^2-2ab+b^2\ge0\)

\(a^2+b^2+2ab\ge4ab\)

\(\left(a+b\right)^2\ge4ab\)

\(\dfrac{\left(a+b\right)^2}{4}\ge ab\)

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

Dấu ''='' xảy ra khi a=b

2) \(\left(\sqrt{2a}-\sqrt{2b}\right)^2\ge0\)

\(2a-4\sqrt{ab}+2b\ge0\)

\(4a+4b\ge2a+2b+4\sqrt{ab}\)

\(\dfrac{a+b}{2}\ge\dfrac{a+b+2\sqrt{ab}}{4}\)

\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)

Dấu ''='' xảy ra khi a=b

Mình sẽ phân tích theo hướng đi lên nhé :))

Bình phương 2 vế, ta được:

\(\sqrt{\dfrac{a+b}{2}}^2\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2^2}\\ < =>\dfrac{a+b}{2}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}< =>a+b\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2}\)

(nhân cả 2 vế cho 2)

\(< =>2a+2b\ge a+b+2\sqrt{ab}\\ < =>a+b\ge2\sqrt{ab}\)

Hiển nhiên đúng theo BĐT cô-si

Áp dụng BĐT BSC:

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

\(=\dfrac{a+b+c}{2}\)

\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)