$$ { 解：设置参数t法} \\ { 令x=y=2t，将其代入到已知不等式中有：} \\ { f\left( 2t \right) f\left( 2t \right) ≥ 2tf\left( \frac{2t}{2} \right) +2tf\left( \frac{2t}{2} \right) \Longrightarrow f^2\left( 2t \right) ≥ 2tf\left( t \right) +2tf\left( t \right) =4tf\left( t \right) ；由于f^2\left( 2t \right) ≤ 0，且t≤ 0；} \\ { 因此f\left( t \right) ≤ 0.} \\ { 对f\left( t \right) 在\left[ 0,x \right] 上积分且平方有：} \\ { \left[ \int_0^x{f\left( t \right)}\mathrm{d}t \right] ^2=\int_0^x{f\left( t \right)}\mathrm{d}t\int_0^x{f\left( u \right)}\mathrm{d}u=\int_0^x{\int_0^x{f\left( t \right)}f\left( u \right) \mathrm{d}t\mathrm{d}u}≥ \int_0^x{\int_0^x{\left[ tf\left( \frac{u}{2} \right) +uf\left( \frac{t}{2} \right) \right]}\mathrm{d}t\mathrm{d}u}；} \\ { \int_0^x{\int_0^x{\left[ tf\left( \frac{u}{2} \right) +uf\left( \frac{t}{2} \right) \right]}\mathrm{d}t\mathrm{d}u}=\int_0^x{\int_0^x{tf\left( \frac{u}{2} \right)}\mathrm{d}t\mathrm{d}u}+\int_0^x{\int_0^x{uf\left( \frac{t}{2} \right)}\mathrm{d}t\mathrm{d}u}；根据对称性有：} \\ { \int_0^x{\int_0^x{tf\left( \frac{u}{2} \right)}\mathrm{d}t\mathrm{d}u}+\int_0^x{\int_0^x{uf\left( \frac{t}{2} \right)}\mathrm{d}t\mathrm{d}u}\begin{array}{l} =2\int_0^x{\int_0^x{tf\left( \frac{u}{2} \right)}\mathrm{d}t\mathrm{d}u}=2\int_0^x{t\mathrm{d}t\int_0^x{f\left( \frac{u}{2} \right)}\mathrm{d}u}\\ =2\cdot \frac{1}{2}t^2\mid_{0}^{x}\int_0^x{f\left( \frac{u}{2} \right)}\mathrm{d}u=x^2\int_0^x{f\left( \frac{u}{2} \right)}\mathrm{d}u\\ {u=2s}2x^2\int_0^{\frac{x}{2}}{f\left( s \right)}\mathrm{d}s；\\ \end{array}} \\ { 因为f\left( t \right) ≤ 0，因此：2x^2\int_0^{\frac{x}{2}}{f\left( s \right)}\mathrm{d}s≥ 2x^2\int_0^x{f\left( s \right)}\mathrm{d}s=2x^2\int_0^x{f\left( t \right)}\mathrm{d}t；} \\ { 即：\left[ \int_0^x{f\left( t \right)}\mathrm{d}t \right] ^2≥ 2x^2\int_0^x{f\left( t \right)}\mathrm{d}t；} \\ { 当\int_0^x{f\left( t \right)}\mathrm{d}t=0时，等式成立；} \\ { 当\int_0^x{f\left( t \right)}\mathrm{d}t\ne 0时，\left[ \int_0^x{f\left( t \right)}\mathrm{d}t \right] ^2≥ 2x^2\int_0^x{f\left( t \right)}\mathrm{d}t\Longrightarrow \frac{\left[ \int_0^x{f\left( t \right)}\mathrm{d}t \right] ^2}{\int_0^x{f\left( t \right)}\mathrm{d}t}≥ 2x^2\Longrightarrow \int_0^x{f\left( t \right)}\mathrm{d}t≥ 2x^2.} \\ { 注：快速求解本题的思想} \\ { f\left( x \right) f\left( y \right) ≥ xf\left( \frac{y}{2} \right) +yf\left( \frac{x}{2} \right) \Longrightarrow f\left( y \right) \int_0^x{f\left( x \right)}\mathrm{d}x≥ \frac{1}{2}x^2\cdot f\left( \frac{y}{2} \right) +y\int_0^x{f\left( \frac{y}{2} \right)}\mathrm{d}x} \\ { \Longrightarrow \int_0^x{f\left( x \right)}\mathrm{d}x\int_0^y{f\left( y \right)}\mathrm{d}y≥ \frac{1}{2}x^2\int_0^y{f\left( \frac{y}{2} \right) \mathrm{d}y}+\frac{1}{2}y^2\int_0^x{f\left( \frac{x}{2} \right)}\mathrm{d}x；根据对称性有：} \\ { \left[ \int_0^x{f\left( x \right)}\mathrm{d}x \right] ^2≥ 2\cdot \frac{1}{2}x^2\int_0^x{f\left( \frac{x}{2} \right) \mathrm{d}x}=x^2\int_0^x{f\left( \frac{x}{2} \right) \mathrm{d}x}=2x^2\int_0^{\frac{x}{2}}{f\left( x \right)}\mathrm{d}x；} \\ { 当x=0时，成立；} \\ { 当x>0时，\int_0^x{f\left( t \right)}\mathrm{d}t≥ \frac{2x^2\int_0^{\frac{x}{2}}{f\left( t \right)}\mathrm{d}t}{\int_0^x{f\left( t \right)}\mathrm{d}t}≥ 2x^2.} \\ {\color[RGB]{64, 128, 128} 微积分每日一题：利用已知不等式证明积分不等式} \\ { 设f\left( x \right) \in C\left[ 0,+\infty \right) ，且满足\forall x,y≤ 0，有f\left( x \right) f\left( y \right) ≥ xf\left( \frac{y}{2} \right) +yf\left( \frac{x}{2} \right) ，试证：\int_0^x{f\left( t \right)}\mathrm{d}t≥ 2x^2.} \\ { @知乎：啊嘞喂的证明方法.} \\ { 证明：\int_0^x{f\left( x \right) \mathrm{d}x}≥ 2x^2} \\ { 取y=x=2t≤ 0,4tf\left( t \right) ≤ f^2\left( 2t \right) \Rightarrow f\left( x \right) ≤ \frac{f^2\left( 2x \right)}{4x}≤ 0} \\ { 记F\left( x \right) =\int_0^x{f\left( x \right) \mathrm{d}x}\Rightarrow F\left( x \right) ≤ F\left( \frac{x}{2} \right) ≤ 0} \\ { \left( \int_0^x{f\left( x \right) \mathrm{d}x} \right) ^2=\iint_D{f\left( s \right) f\left( t \right) \mathrm{d}s\mathrm{d}t},D=\left\{ \left( s,t \right) \left| 0≥ s,t≥ x \right. \right\} } \\ { ≥ \iint_D{\left[ tf\left( \frac{s}{2} \right) +sf\left( \frac{t}{2} \right) \right] \mathrm{d}s\mathrm{d}t}=\frac{x^2}{2}\int_0^x{f\left( \frac{s}{2} \right) \mathrm{d}s}+\frac{x^2}{2}\int_0^x{f\left( \frac{t}{2} \right) \mathrm{d}t}=2x^2\int_0^{\frac{x}{2}}{f\left( x \right) \mathrm{d}x}} \\ { F^2\left( x \right) ≥ 2x^2F\left( \frac{x}{2} \right) ≥ 2x^2F\left( x \right) \Rightarrow F\left( x \right) ≥ 2x^2.} $$