Prove that for any real $ x $ and $ y $ the inequality$$x^2 \sqrt {1+32y^2} + y^2 \sqrt {1+32x^2} \geq 4xy (x+y+\frac{\sqrt{2}}{4})$$$$x^2 \sqrt {1+64y^2} + y^2 \sqrt {1+64x^2} \geq 4xy (x+y+\frac{\sqrt{3}}{4})$$ parmenides51 wrote: Prove that for any real $ x $ and $ y $ the inequality $x^2 \sqrt {1+2y^2} + y^2 \sqrt {1+2x^2} \geq xy (x+y+\sqrt{2})$ . Proof of Zhangyunhua(张云华)：

Prove that for any real $ x $ and $ y $ the inequality $$x^2 \sqrt {1+4y^2} + y^2 \sqrt {1+4x^2} \geq xy (x+y+\sqrt{3})$$$$x^2 \sqrt {1+y^2} + y^2 \sqrt {1+x^2} \geq \frac{1}{2}xy (x+y+2\sqrt{3})$$$$x^2 \sqrt {1+8y^2} + y^2 \sqrt {1+8x^2} \geq 2xy (x+y+\frac{\sqrt{2}}{2})$$$$x^2 \sqrt {1+16y^2} + y^2 \sqrt {1+16x^2} \geq 2xy (x+y+\frac{\sqrt{3}}{2})$$

lemma:√2(a+b)≥ √a + √b (the proof is based on AM≥GM) From here we can see that √1+2y²≥√y² + √1/2=y+ √1/2 We get x²y+ x²/√2 + y²x+ y²/√2≥x²y+xy²+ xy√2 x²/√2 + y²/√2≥xy√2 ; and this can be proven by AM≥GM: a²+b²≥2ab.