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

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

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

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

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

a.

Bình phương 2 vế, BĐT cần chứng minh trở thành:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)

Ta có:

\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)

Tương tự cộng lại:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b.

\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)

Nên ta chỉ cần chứng minh:

\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)

\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)

Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)

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?

\(\text{Ta có }:\left(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\right)^2\\ =x^2+y^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}+z^2+t^2\)

Áp dụng định lí bu-nhi-a-cốp-xki:

\(\Rightarrow2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge2\sqrt{\left(xz+yt\right)^2}=2xz+2yt\\ \Rightarrow\left(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\right)^2\\ \ge x^2+y^2+2xz+2yt+z^2+t^2\\ =x^2+2xz+z^2+y^2+2yt+t^2\\ =\left(x+z\right)^2+\left(y+t\right)^2\\ \Rightarrow\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{y}=\frac{z}{t}\)

Áp dụng BDT trên

\(\Rightarrow\sqrt{a^2+b^2-\sqrt{3}ab}+\sqrt{b^2+c^2-bc}\\ =\sqrt{\frac{3}{4}a^2-\sqrt{3}ab+b^2+\frac{1}{4}a^2}+\sqrt{b^2-bc+\frac{1}{4}c^2+\frac{3}{4}c^2}\\ =\sqrt{\left(\frac{\sqrt{3}}{2}a-b\right)^2+\frac{1}{4}a^2}+\sqrt{\left(b-\frac{1}{2}c\right)^2+\frac{3}{4}c^2}\\ \ge\sqrt{\left(\frac{\sqrt{3}}{2}a-b+b-\frac{1}{2}c\right)^2+\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}c\right)^2}\\ =\sqrt{\left(\frac{\sqrt{3}}{2}a-\frac{1}{2}c\right)^2+\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}c\right)^2}\\ =\sqrt{\frac{3}{4}a^2-\frac{\sqrt{3}}{2}ac+\frac{1}{4}c^2+\frac{1}{4}a^2+\frac{\sqrt{3}}{2}ac+\frac{3}{4}c^2}\\ \\ =\sqrt{a^2+c^2}\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{\frac{\sqrt{3}}{2}a-b}{\frac{1}{2}a}=\frac{b-\frac{1}{2}c}{\frac{\sqrt{3}}{2}c}\)

\(\Leftrightarrow\frac{\sqrt{3}a-2b}{a}=\frac{2b-c}{\sqrt{3}c}\\ \Leftrightarrow\sqrt{3}c\left(\sqrt{3}a-2b\right)=a\left(2b-c\right)\\ \Leftrightarrow3ac-2\sqrt{3}bc=2ab-ac\\ \Leftrightarrow4ac-2\sqrt{3}bc-2ab=0\)

Áp dụng BĐT: \(x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\)

\(\Rightarrow VT\ge\sqrt{\frac{1}{2}\left(a+b\right)^2}+\sqrt{\frac{1}{2}\left(b+c\right)^2}+\sqrt{\frac{1}{2}\left(c+a\right)^2}\)

\(VT\ge\frac{1}{\sqrt{2}}\left(2a+2b+2c\right)=\sqrt{2}\)

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

Ta sẽ chứng minh:

\(\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge\sqrt{\left(a+b\right)^2+\left(x+y\right)^2}\)

Thật vậy, bình phương 2 vế, BĐT tương đương:

\(a^2+x^2+b^2+y^2+2\sqrt{a^2b^2+x^2y^2+a^2y^2+b^2x^2}\ge a^2+b^2+x^2+y^2+2ab+2xy\)

\(\Leftrightarrow\sqrt{a^2b^2+x^2y^2+a^2y^2+b^2x^2}\ge ab+xy\)

\(\Leftrightarrow a^2b^2+x^2y^2+a^2y^2+b^2x^2\ge a^2b^2+x^2y^2+2abxy\)

\(\Leftrightarrow a^2y^2+b^2x^2-2abxy\ge0\)

\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) (luôn đúng)

Áp dụng:

\(VT=\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\)

\(VT\ge\sqrt{\left(a+b\right)^2+\left(x+y\right)^2}+\sqrt{c^2+z^2}\ge\sqrt{\left(a+b+c\right)^2+\left(x+y+z\right)^2}\) (đpcm)

Đề bài sai

Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\Rightarrow xyz=1\)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(P=\dfrac{1}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}+\dfrac{1}{\left(y^2+z^2\right)+\left(z^2+1\right)+2}+\dfrac{1}{\left(z^2+x^2\right)+\left(x^2+1\right)+2}\)

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{xz\left(xy+y+1\right)}+\dfrac{x}{x\left(yz+z+1\right)}+\dfrac{1}{zx+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x.xyz+xyz+xz}+\dfrac{x}{xyz+xz+1}+\dfrac{1}{xz+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x+1+xz}+\dfrac{x}{1+xz+1}+\dfrac{1}{xz+x+1}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)