\(\left\{{}\begin{matrix}x;y>0\\x+y=1\end{matrix}\right.\)\(\Rightarrow0< xy=t\le\dfrac{1}{4}\)

\(x^4+y^4=\left(1-2t\right)^2-2t\)

\(8\left(x^4+y^4\right)+\dfrac{1}{xy}\ge5\Leftrightarrow A=8\left[\left(1-2t\right)^2-2t\right]+\dfrac{1}{t}-5\ge0\)

\(\Leftrightarrow16t^2-32t+\dfrac{1}{t}+3\ge0\)\(\Leftrightarrow16t^3-32t^2+3t+1\ge0\)

<=>\(16t^3-4t^2-28t^2+7t-4t+1\ge0\)

\(\Leftrightarrow4t^2\left(4t-1\right)-7t\left(4t-1\right)-\left(4t-1\right)\ge0\)

\(\Leftrightarrow\left(4t-1\right)\left(4t^2-7t-1\right)\ge0\)

\(\Leftrightarrow B=\left(4t-1\right)\left(8t-7-\sqrt{65}\right)\left(8t-7+\sqrt{65}\right)\ge0\)

\(0< t\le\dfrac{1}{4}\Rightarrow\)\(\left\{{}\begin{matrix}4t-1\le0\\8t-7+\sqrt{65}>0\\8t-7-\sqrt{5}< 0\end{matrix}\right.\) \(\Rightarrow B\ge0\)

mọi phép biến đổi <=> => dpcm

Sử dụng BĐT Cauchy-Schwarz nhiều lần, cộng với BĐT phụ \(\dfrac{1}{xy}\ge\dfrac{4}{\left(x+y\right)^2}\), ta có:

\(8\left(x^4+y^4\right)+\dfrac{1}{xy}\ge\dfrac{8\left(x^2+y^2\right)^2}{2}+\dfrac{4}{\left(x+y\right)^2}=4\left(x^2+y^2\right)^2+4\ge4\left[\dfrac{\left(x+y\right)^2}{2}\right]^2+4=5\)

Đẳng thức xảy ra khi \(x=y=\dfrac{1}{2}\)

 có bđt: a²+b² ≥ (a+b)²/2 (*) 
(*) <=> 2a²+2b² ≥ a²+b²+2ab <=> a²+b²-2ab ≥ 0 <=> (a-b)² ≥ 0 bđt đúng, dấu "=" khi a = b 
- - - 
ad (*) 2 lần liên tiếp: 
x^4 + y^4 ≥ (x²+y²)²/2 ≥ [(x+y)²/2]²/2 = (x+y)^4 /8 = 1/8 
=> 8(x^4 + y^4) ≥ 1 (*) 

mặt khác, có bđt: (x-y)² ≥ 0 <=> x²+y² ≥ 2xy <=> x²+y²+2xy ≥ 4xy <=> (x+y)² ≥ 4xy 
=> 1/xy ≥ 4/(x+y)² = 4 (**) 

(*) + (**): 8(x^4 + y^4) + 1/xy ≥ 1+4 = 5 (đpcm) dấu "=" khi x = y = 1/2 

\(A=8\left(x^4+y^4\right)+\frac{1}{4xy}+\frac{1}{4xy}+\frac{1}{2xy}\ge8\left(x^4+y^4\right)+\frac{1}{2\left(x^2+y^2\right)}+\frac{1}{2\left(x^2+y^2\right)}+\frac{1}{2xy}\)

\(\Rightarrow A\ge8\left(x^4+y^4\right)+\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}+\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}+\frac{1}{2\left(\frac{x+y}{2}\right)^2}\)

\(\Rightarrow A\ge3\sqrt[3]{8\left(x^4+y^4\right)\cdot\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}\cdot\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}}+\frac{1}{2\cdot\frac{1}{4}}=3+2=5\)

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

Là sao ko hiểu đề

Áp dụng BĐT Cô-si :

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

Do đó BĐT cần chứng minh \(\Leftrightarrow8\left(x^4+y^4\right)+4\ge5\)

Ta cần chứng minh BĐT sau là đủ : \(8\left(x^4+y^4\right)\ge1\)

Thật vậy: Áp dụng BĐT Cô-si :

\(x^4+\frac{1}{16}\ge\frac{x^2}{2};y^4+\frac{1}{16}\ge\frac{y^2}{2}\)

Cộng vế : \(x^4+y^4+\frac{1}{8}\ge\frac{x^2+y^2}{2}\ge\frac{\frac{\left(x+y\right)^2}{2}}{2}\ge\frac{\frac{1}{2}}{2}=\frac{1}{4}\)

\(\Leftrightarrow x^4+y^4\ge\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\)

\(\Leftrightarrow8\left(x^4+y^4\right)\ge1\)

Ta có đpcm.

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

\(x+y=1\ge2\sqrt{xy}\Leftrightarrow xy\le\frac{1}{4}\)

\(A=8\left(x^4+y^4\right)+\frac{1}{xy}\ge16x^2y^2+\frac{1}{xy}=16x^2y^2+\frac{1}{4xy}+\frac{1}{4xy}+\frac{1}{2xy}\ge3\sqrt[3]{16x^2y^2.\frac{1}{4xy}.\frac{1}{4xy}}+\frac{1}{2.\frac{1}{4}}=5\)

Dâu ' = ' xảy ra khi  x =y = 1/2