\(y\ge xy+1\ge2\sqrt{xy}\Rightarrow\sqrt{\dfrac{y}{x}}\ge2\Rightarrow\dfrac{y}{x}\ge4\)

\(Q=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{\dfrac{y}{x}+\left(\dfrac{y}{x}\right)^2}\)

Đặt \(\dfrac{y}{x}=a\ge4\)

\(Q=\dfrac{2a^2-2a+1}{a^2+a}=\dfrac{2a^2-2a+1}{a^2+a}-\dfrac{5}{4}+\dfrac{5}{4}=\dfrac{\left(a-4\right)\left(3a-1\right)}{4\left(a^2+1\right)}+\dfrac{5}{4}\ge\dfrac{5}{4}\)

\(Q_{min}=\dfrac{5}{4}\) khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\Rightarrow\dfrac{y}{x}\ge4\)

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

Đặt \(\dfrac{y}{x}=a\ge4\Rightarrow P=\dfrac{2a^2-2a+1}{a+1}=2a-4+\dfrac{5}{a+1}\)

\(P=\dfrac{a+1}{5}+\dfrac{5}{a+1}+\dfrac{9}{5}.a-\dfrac{21}{5}\ge2\sqrt{\dfrac{5\left(a+1\right)}{5\left(a+1\right)}}+\dfrac{9}{5}.4-\dfrac{21}{5}=5\)

Dấu "=" xảy ra khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Giúp mình câu này với ah. 

Áp dụng bđt Schwarz ta có:

\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).

Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

1) \(\left\{{}\begin{matrix}a^3+b^3+c^3=3abc\\a+b+c\ne0\end{matrix}\right.\)  \(\left(a;b;c\in R\right)\)

Ta có :

\(a^3+b^3+c^3\ge3abc\) (Bất đẳng thức Cauchy)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\left(a^3+b^3+c^3=3abc\right)\)

Thay \(a=b=c\) vào \(P=\dfrac{a^2+2b^2+3c^2}{3a^2+2b^2+c^2}\) ta được

\(\Leftrightarrow P=\dfrac{6a^2}{6a^2}=1\)

\(3^x=y^2+2y\left(x;y>0\right)\)

\(\Leftrightarrow3^x+1=y^2+2y+1\)

\(\Leftrightarrow3^x+1=\left(y+1\right)^2\left(1\right)\)

- Với \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

\(pt\left(1\right)\Leftrightarrow3^0+1=\left(0+1\right)^2\Leftrightarrow2=1\left(vô.lý\right)\)

- Với \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)  

\(pt\left(1\right)\Leftrightarrow3^1+1=\left(1+1\right)^2=4\left(luôn.luôn.đúng\right)\)

- Với \(x>1;y>1\)

\(\left(y+1\right)^2\) là 1 số chính phương

\(3^x+1=\overline{.....1}+1=\overline{.....2}\) không phải là số chính phương

\(\Rightarrow\left(1\right)\) không thỏa với \(x>1;y>1\)

Vậy với \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) thỏa mãn đề bài

\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)

\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)

\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)

12. Ta có \(ab\le\frac{a^2+b^2}{2}\)

=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)

Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)

=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)

=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)

Khi đó 

\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)

Dấu bằng xảy ra khi a=b=c=1

Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1

13.  Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)

=> \(1\ge\frac{9}{a+b+c+3}\)

=> \(a+b+c\ge6\)

Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)

Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)

Cộng 3 BT trên ta có

\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)

Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)

=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)

Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)

<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)

<=> \(a^2+b^2\ge2ab\)(luôn đúng )

=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)

=> \(P\ge2\)

Vậy \(MinP=2\)khi a=b=c=2

Lưu ý : Chỗ .... là tương tự 

14.

Ta có \(x^3+y^3+z^3-3xyz=2\)

=> \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=2\)

Đặt \(x+y+z=a,x^2+y^2+z^2-xy-yz-xz=b\left(a,b\ge0\right)\)

=> \(\hept{\begin{cases}ab=2\\P=\frac{1}{2}a^2+4b\end{cases}}\)

Áp dụng bất đẳng thức cosi cho 3 số ta có 

\(\frac{1}{2}a^2+2b+2b\ge3\sqrt[3]{2a^2b^2}=6\)

Dấu bằng xảy ra khi \(\hept{\begin{cases}\frac{1}{2}a^2=2b\\ab=2\end{cases}}\)=> \(\hept{\begin{cases}a=2\\b=1\end{cases}}\)thỏa mãn ĐK

Vậy MinP=6 khi a=2, b=1