Note that, $a(c^2-1)=b(b^2+c^2)$ implies that $(a-b)c^2=a+b^3$. Now, our goal is to show, $$ d^2(a\sqrt{1-d^2}+b^2\sqrt{1+d^2})^2 \leqslant \frac{(a+b)^2 c^2}{4}. $$Observe that, by Cauchy-Schwarz, $$ (a+b^3)(a(1-d^2)+b(1+d^2))\geqslant (a\sqrt{1-d^2}+b^2\sqrt{1+d^2})^2. $$Hence, $$ d^2(a\sqrt{1-d^2}+b^2\sqrt{1+d^2})^2\leqslant d^2(a+b^3)(a(1-d^2)+b(1+d^2))=d^2(a-b)c^2(a+b -(a-b)d^2). $$Now since $$ d^2(a-b)(a+b-(a-b)d^2)\leqslant \left(\frac{a+b}{2}\right)^2, $$we get an upper bound of $\frac{c^2(a+b)^2}{4}$, as claimed.

izat wrote: For positive real numbers $a, b, c, d,$ satisfying the following conditions: $a(c^2 - 1)=b(b^2+c^2)$ and $d \leq 1$, prove that : $d(a \sqrt{1-d^2} + b^2 \sqrt{1+d^2}) \leq \frac{(a+b)c}{2}$ anhduy98: $$LHS=d\left( \sqrt{a}\sqrt{a(1-d^2)}+\sqrt{b^3}\sqrt{b(1+d^2)}\right) \le d \sqrt{a+b^3}\sqrt{a(1-d^2)+b(1+d^2)}$$$$\Rightarrow LHS \le d \sqrt{(a-b)c^2}\sqrt{a+b-(a-b)d^2}=c \sqrt{(a-b)d^2}\sqrt{a+b-(a-b)d^2} \le \frac{(a+b)c}{2}.$$

grupyorum wrote: Note that, $a(c^2-1)=b(b^2+c^2)$ implies that $(a-b)c^2=a+b^3$. Now, our goal is to show, $$ d^2(a\sqrt{1-d^2}+b^2\sqrt{1+d^2})^2 \leqslant \frac{(a+b)^2 c^2}{4}. $$Observe that, by Cauchy-Schwarz, $$ (a+b^3)(a(1-d^2)+b(1+d^2))\geqslant (a\sqrt{1-d^2}+b^2\sqrt{1+d^2})^2. $$Hence, $$ d^2(a\sqrt{1-d^2}+b^2\sqrt{1+d^2})^2\leqslant d^2(a+b^3)(a(1-d^2)+b(1+d^2))=d^2(a-b)c^2(a+b -(a-b)d^2). $$Now since $$ d^2(a-b)(a+b-(a-b)d^2)\leqslant \left(\frac{a+b}{2}\right)^2, $$we get an upper bound of $\frac{c^2(a+b)^2}{4}$, as claimed. very nice