Có: \(a+b+c=1\Leftrightarrow\left(a+b+c\right)^2=1\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}\)

\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\) (do \(\left(a+b+c\right)^2=a^2+b^2+c^2=1\))

\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)

\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

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

\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)

\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)

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

a) 3.(ab+bc+ac)≤a²+b²+c²+2ab+2bc+2ac

<=> \(a^2+b^2+c^2-ab-bc-ac\ge0\)

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

<=> (a-b)²+(b-c)²+(a-c)²≥0 ( luon dung voi moi a,b,c)

b) ap dung ket qua tren va vế sau bn xem bài giải của mk ở trên

Xét hiệu VT - VP

\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ab+b^2}+\dfrac{c+a}{ab+c^2}-\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}=\dfrac{a^2+ab-bc-a^2}{a\left(bc+a^2\right)}+\dfrac{b^2+bc-ac-b^2}{b\left(ac+b^2\right)}+\dfrac{c^2+ac-ab-c^2}{c\left(ab+c^2\right)}=\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}\)

Do a,b,c bình đẳng nên giả sử a\(\ge\)b\(\ge\)c, khi đó \(b\left(a-c\right)\)\(\ge\)0, c(b-a)\(\le\)0, a(c-b)\(\le\)0

\(a^3\ge b^3\ge c^3=>abc+a^3\ge abc+b^3\ge abc+c^3\)=>\(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}\le\dfrac{b\left(a-c\right)}{b\left(ac+b^2\right)}\)

=> VT -VP \(\le\) \(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}=\dfrac{ab-ac}{b\left(ac+b^2\right)}+\dfrac{ac-ab}{c\left(ab+c^2\right)}=\dfrac{a\left(b-c\right)}{b\left(ac+b^2\right)}-\dfrac{a\left(b-c\right)}{c\left(ab+c^2\right)}\)

mà \(\dfrac{1}{b\left(ac+b^2\right)}\le\dfrac{1}{c\left(ab+c^2\right)}\) nên VT-VP <0 đpcm

Ta viết bất đẳng thức đã cho lại thành

\(\sum\left[\dfrac{1}{c}-\dfrac{\left(a+b+2c\right)}{2\left(ab+c^2\right)}\right]\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{2\prod\left(ab+c^2\right)}\)

\(\Leftrightarrow\sum\dfrac{c\left(a^2+ab+b^2\right)\left(a-b\right)^2}{ab\left(a^2+bc\right)\left(b^2+ca\right)}\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}\)

Hay \(S_a\left(b-c\right)^2+S_b\left(c-a\right)^2+S_c\left(a-b\right)^2\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}\quad\left(1\right)\)

Vậy $VT\geq 0$ và $S_a+S_b\ge 0;S_b+S_c\ge 0.$ Nếu \(a\ge b\ge c\rightarrow VT\ge0\ge VP,\) ta chỉ xét \(a\le b\le c.\)

\(\left(1\right)\Leftrightarrow\left(S_a+S_b\right)\left(b-c\right)^2+\left(S_b+S_c\right)\left(a-b\right)^2\ge\left[\dfrac{\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}-2S_b\right]\left(a-b\right)\left(b-c\right)\)

Đặt \(c=a+x+y,b=a+x\Rightarrow x=b-a;y=c-b\left(x,y\ge0\right)\) thay vào rút gọn các thứ là đpcm.

P/s: Cách này khá trâu nhưng chịu thôi, bài này mình nghĩ khá chặt.

a) Sai đề.

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

b) Sai đề.

\(\dfrac{a\sqrt[]{b}+b\sqrt[]{a}}{\sqrt[]{ab}}:\dfrac{1}{\sqrt[]{a}-\sqrt[]{b}}=\dfrac{\sqrt[]{ab}\left(\sqrt[]{a}+\sqrt[]{b}\right)}{\sqrt[]{ab}}.\left(\sqrt[]{a}-\sqrt[]{b}\right)=a-b\)

c) \(\left(\dfrac{\sqrt{x}+\sqrt[]{y}}{\sqrt[]{x}-\sqrt[]{y}}-\dfrac{\sqrt[]{x}-\sqrt[]{y}}{\sqrt[]{x}+\sqrt[]{y}}\right):\dfrac{\sqrt[]{xy}}{x-y}\)

\(=\dfrac{\left(\sqrt[]{x}+\sqrt[]{y}\right)^2-\left(\sqrt[]{x}-\sqrt[]{y}\right)^2}{\left(\sqrt[]{x}-\sqrt[]{y}\right)\left(\sqrt[]{x}+\sqrt[]{y}\right)}.\dfrac{x-y}{\sqrt[]{xy}}=\dfrac{4\sqrt[]{xy}}{x-y}.\dfrac{x-y}{\sqrt[]{xy}}=4\)

Lời giải:

a) Vì $abc=1$ nên ta có:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=\frac{ac}{abc.+ac+c}+\frac{b.ac}{bc.ac+b.ac+ac}+\frac{c}{ac+c+1}\)

\(=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}=\frac{ac+1+c}{ac+c+1}=1\)

(đpcm)

b)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow \left\{\begin{matrix} x=ka\\ y=kb\\ z=kc\end{matrix}\right.\)

\(x+y+z=ka+kb+kc=k(a+b+c)=k\)

\(x^2+y^2+z^2=k^2a^2+k^2b^2+k^2c^2=k^2(a^2+b^2+c^2)=k^2\)

\(\Rightarrow A=xy+yz+xz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{k^2-k^2}{2}=0\)

a) Từ giả thiết : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)

\(\Rightarrow2ab\text{=}2bc+2ca\)

\(\Rightarrow2ab-2bc-2ca\text{=}0\)

Ta xét : \(\left(a+b-c\right)^2\text{=}a^2+b^2+c^2+2ab-2bc-2ca\)

\(\text{=}a^2+b^2+c^2\)

Do đó : \(A\text{=}\sqrt{a^2+b^2+c^2}\text{=}\sqrt{\left(a+b-c\right)^2}\)

\(\Rightarrow A\text{=}a+b-c\)

Vì a;b;c là các số hữu tỉ suy ra : đpcm

b) Đặt : \(a\text{=}\dfrac{1}{x-y};b\text{=}\dfrac{1}{y-x};c\text{=}\dfrac{1}{z-x}\)

Do đó : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)

Ta có : \(B\text{=}\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

Từ đây ta thấy giống phần a nên :

\(B\text{=}a+b-c\)

\(B\text{=}\dfrac{1}{x-y}+\dfrac{1}{y-z}-\dfrac{1}{z-x}\)

Suy ra : đpcm.

Mình bổ sung đề phần b cần phải có điều kiện của x;y;z nha bạn.