Ta có : \(VP=\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{yx}}=2\)

Vậy \(Q_{min}=2\)với \(x=y\)

mình không chắc về phân bđt này lắm

Đặt x=a, \(\frac{1}{y}=b\)\(\Rightarrow a+b\le1\)

Ta có: \(Q=ab+\frac{1}{ab}=16ab+\frac{1}{ab}-15ab\ge2\sqrt{\frac{16ab}{ab}}-\frac{15.\left(a+b\right)^2}{4}=8-\frac{15.1}{4}=\frac{17}{4}\)

Dấu "=" xảy ra khi a=b=\(\frac{1}{2}\)hay \(x=\frac{1}{2},y=2\)

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By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

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

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

Đặt: \(\frac{1}{y}=t\)> 0

Ta có: \(x+t\le1\)

\(P=\frac{xt}{2}+\frac{1}{xt}=\frac{xt}{2}+\frac{1}{32xt}+\frac{31}{32xt}\ge2\sqrt{\frac{xt}{2}.\frac{1}{32xt}}+\frac{31}{\frac{32\left(x+t\right)^2}{4}}=\frac{33}{8}\)

Dấu "=" xảy ra <=> x = t = 1/2 hay x = 1/2 và y = 2 

Vậy GTNN của P = 33/8 đạt tại x =1/2 và y =2 .

\(A\ge\frac{1}{2}\left(x+y\right)^2+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)^2+4\)

\(A\ge\frac{1}{2}\left(x+y\right)^2+\frac{8}{\left(x+y\right)^2}+4=\frac{1}{2}\left(x+y\right)^2+\frac{1}{2\left(x+y\right)^2}+\frac{15}{2\left(x+y\right)^2}+4\)

\(A\ge2\sqrt{\frac{\left(x+y\right)^2}{4\left(x+y\right)^2}}+\frac{15}{2.1}+4=\frac{25}{2}\)

\(A_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)

Trước tiên chứng minh:

\(a^4+b^4\ge a^3b+ab^3\)

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

\(\Rightarrow2\left(a^4+b^4\right)\ge a^4+b^4+a^3b+ab^3=\left(a+b\right)\left(a^3+b^3\right)\)

Áp dụng bài toán được

\(P=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^3+x^3}\)

\(\ge\frac{1}{2}\left(x+y+y+z+z+x\right)=x+z+y=2018\)

Bổ đề: \(2xy\le x^2+y^2\)

\(A=\frac{1}{x^2+y^2}+\frac{2}{xy}=\frac{1}{x^2+y^2}+\frac{4}{2xy}\ge\frac{1}{x^2+y^2}+\frac{4}{x^2+y^2}=\frac{5}{x^2+y^2}\ge5\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{\sqrt{2}}\)

Bài làm:

Ta có: \(C=\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge\frac{\left(1+2+3\right)^2}{x+y+z}\ge\frac{6^2}{1}=36\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=\frac{1}{13}\\y=\frac{4}{13}\\z=\frac{9}{13}\end{cases}}\)

Bài này là áp dụng bđt Cauchy-Schwaz nha bạn.

Mk làm lại nha, bài kia mk lm sai!

Ta có: \(C=\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge\frac{36}{x+y+z}\ge\frac{36}{1}=36\)

Dấu "=" xảy ra khi: \(\frac{1}{x}=\frac{2}{y}=\frac{3}{z}\Rightarrow\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=\frac{1}{3}\\z=\frac{1}{2}\end{cases}}\)

\(\frac{y}{x}+\frac{x}{y}\ge2\left(Cauchy\right)\Rightarrow Min=2\Leftrightarrow x=y\)